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Full text of "Advanced geometry for high schools : synthetic and analytical"

ADVANCED GEOMETRY 
FOE HIGH SGHOOLS 

SYNTHETIC -ANALYTICAL 

M3DOUGALL. 



REVISED EDITION 




Digitized by the Internet Archive 

in 2009 with funding from 

Ontario Council of University Libraries 



http://www.archive.Org/details/1 91 9advancedgeom00mcdo 



ADVANCED GEOMETRY 

FOR 

HIGH SCHOOLS 

SYNTHETIC AND ANALYTICAL 



PART I.— SYNTHETIC GEOMETRY 
PART II.— ANAL YTICAL GEOMETR Y 



A. H. McDOUGALL, B.A., LL.D. 

Principal Ottawa Collegiate Institute 



SECOND EDITION 

REVISED AXD AMPLIFIED 



TORONTO 

THE COPP CLARK COMPANY, LIMITED 



Copyright, Canada, 1919, by The Copp Clark Company, Limited, 
Toronto, Ontario 



PREFACE TO THE SECOND EDITION 



In this edition changes have been made in Part I. on the 
suggestion of teachers who have kindly pointed out difficulties 
experienced by their pupils in using the original edition. The 
exercises have been re-arranged and, in general, divided into 
two sections. Those marked (a) are intended for a first 
reading, and are sufficient for candidates for Entrance to 
the Faculties of Education, while the sections marked (b) 
are more suitable for candidates for honours and scholarships. 

Demonstrations have been re-written, diagrams added, 
and notes given suggesting constructions or proofs. Some 
typical solutions illustrate methods to be used in maxima and 
minima. 

Tn many cases, e.g., loci, the difficulties have been di- 
minished. Certain exercises in the First Edition were stated 
as problems. Under the given conditions the students were 
asked to find the required solution. These exercises are here 
changed to theorems. With the same data the conclusion is 
stated and the proof only is to be discovered. 

In both parts considerable additions suitable for honour 
candidates have been made to the miscellaneous exercises. 

Acknowledgments for valuable assistance received from 
him are due to Mr. I. T. Norris, Mathematical Master of the 
Ottawa Collegiate Institute, and to the teachers who have 
given an encouraging approval of the book as well as assist- 
ance in making it more useful. 

Ottawa, February, 1919. 



CONTENTS 



Chapter I 

l'AQB 

Theorems of Menelaus and Ceva 1 

The Nine-Point Circle 11 

Simpson's Line 13 

Areas op Rectangles 18 

Radical Axis 23 

Chapter II 

Medial Section 30 

Miscellaneous Theorems 37 

Similar and Similarly Situated Polygons ... 39 

Chapter III 

Harmonic Ranges and Pencils 43 

The Complete Quadrilateral 49 

Poles and Polars 50 

Maxima and Minima 60 

Miscellaneous Exercises 

Loci 68 

Theorems • '3 

Problems • °* 



Part I.— SYNTHETIC GEOMETRY 



SYNTHETIC GEOMETRY 



CHAPTER I 

Theorems of Menelaus and Ceva 

1. Menelaus' Theorem:— If a transversal cut the 
sides, or the sides produced, of a triangle, the 
product of one set of alternate segments taken in 
circular order is equal to the product of the other set. 

(Note. — The transversal must cut two sides and the third 
side produced, or cut all three produced.} 




C D^ 

Fiq. 1. Fia. 2. 

A transversal cuts the sides BC, CA, AB of the A 
ABC in the points D, E, F respectively, then 
AF . BD . CE = FB . DC . EA. 
Draw AX, BY, CZ j_ to the transversal. 
From similar As : — ^Fx, P F Y 
AF = AX 
FB BY' 

ss-s 

a CE CZ 

cEz.,*FX and ^A = ^ 

By multiplication 



A> x ^X 21- 1 K£*H? 

FB X DC X EA - i ' 



and /. AF . BD . CE = FB . DC . EA. 

(Historical Note.— Menelaus, a Greek, lived in Alexandria, Egypt, about 98 A.D.) 

1 



2 SYNTHETIC GEOMETRY 

2. Converse of Menelaus' Theorem :— If, in A abc 
on two of the sides bc, CA, ab and on the third 
produced, or if on all three produced, points D, E, F 
respectively be taken so that AF . BD . CE= FB. DC. EA, 
the points D, E, F are collinear. 





Join EF, and produce EF to cut BC at Q. 
V FEG is a st. line, 
.'. AF . BG . CE = FB . GC . EA. (§ I.) 

But AF . BD . CE = FB . DC . EA, by hypothesis. 



-,. -j. BG 

dividing, — — 
&' BD 



BG 



GC. 

DC' 



BD 
DC* 



or, by alternation, QC QC 

/. G coincides with D. (O.H.S. Geometry, § 121.) 
.". D. E, F are collinear. 



THEOREMS OF MENELAUS AND CEVA 6 

3. Ceva's Theorem:— If from the vertices of a 
triangle concurrent straight lines be drawn to cut 
the opposite sides, the product of one set of alternate 
segments taken in circular order is equal to the 
product of the other set. 

(Note. — D, E and F must be on the three sides, or on 
one side and on the other two produced.) 




AO, BO, CO drawn from the vertices of A ABC 
cut BC, CA, AB, at D, E, F respectively, then 
AF . BD . CE = FB . DC . EA. 

FOC is a transversal of A ABD, 

.-. AF . BC . DO = FB . CD . OA. (§ 1.) 

BOE is a transversal of A ADC, 
/. AO . DB. CE = OD . BC . EA. 

By multiplication, and division by DO, OA and BC, 
AF . BD . CE = FB . DC . EA. 

(For another proof of this theorem see O.H. S. 
Geometry, § 122, Exercises 12 and 14.) 

(Historical Not*.— Giovanni Ceva, an Italian engineer, died in 1734 A.D.) 



4 SYNTHETIC GEOMETRY 

4. Converse of Ceva's Theorem :— If, in A abc, 
on the three sides BC, CA, AB, or if on one of these 
sides and on the other two produced, points D, E, F 
respectively be taken so that AF . bd . CE = FB . DC . EA, 
the lines AD, be, cf are concurrent. 




Fia. 7. Fig. 8. 

Draw BE, CF and let them cut at O. Join AO and 
let it cut BC at G. 

•/ AG, BE, CF are concurrent, 

.'. AF . BG . CE = FB . GC . EA, (§ 3.) 

But AF . BD . CE - FB . DC . EA, by hypothesis. 

,. .,. BG GC. 

.*. , dividing, =r= = =—? , 

"' o> BD DC 

Dr, by alternation, ^ = ^. 

.*. G coincides with D. 

.*. AD, BE, CF are concurrent. 



THEOREMS OF MENELAUS AND CEVA 5 

(J. The perpendiculars from the vertices of a 
triangle to the opposite sides are concurrent. 




c B 




In A ABC, draw AX ± BC, BY ± CA, CZ J. AB. 
To prove that AX, BY, CZ are concurrent. 



A AZC III A AYB, 
AZ _ CA 
YA ~ AB' 

tii^ AB 
arl ^' ZB = BC ; 
^Y _ BC 
XC CA" 



and 



6,'sfiZ* 



% 



AZ 
YA 



BX 
ZB 



CY 
XC 



CA 
AB 



AB BC _ , 
BC " CA 



AZ . BX . CY = ZB . XC . YA 

:., by § 4, AX, BY, CZ are concurrent. 

(@/>The point where the J_s from the vertices of a 
A to the opposite sides intersect is called the ortho- 
centre of the A. The A formed by joining the feet 
of these J_s, X, Y, Z in Figures 9 or 10, is called the 
orthocentric, or pedal, A. 



6 SYNTHETIC GEOMETRY 

7.— Exercises 
(a) 

1. Show, from the converse of Ceva's Theorem, that the 
medians of a A are concurrent. 

2. In A ABC, the bisectors of the L s A, B, C cut BC, 

be 

CA, AB at D, E, F respectively. Show that AF = r - 

a + b 

If a = 25, b = 35, c = 20, show that AF . BD . CE = 
2062 T Vr 

3 Show, from the converse of Ceva's Theorem, that the 
bisectors of the /.s of a A are concurrent. 

4. In A ABC, the bisector of the interior L at A and 

of the exterior Zs at B, C cut BC, CA, AB at D, E, F 

be 
respectively. Show that AF = r 

If a = 44, b = 33, c = 22, show that AF . BD . CE = 
76665 |. 

>5 Show, from the converse of Ceva's Theorem, that the 
bisector of the Z at one vertex of a A and the bisectors 
of the exterior Z_s at the other two vertices are concurrent. 
6. In A ABC, AX, BY, CZ the ±a to BC, CA, AB 
intersect at O. Show that : — 

(a) rect. AO . OX = rect. BO . OY = rect. CO . OZ ; 

(b) rect. AB . AZ = rect. AO . AX = rect. AC . AY; 

\c) if AX meet the circumscribed circle of A ABC at K, 
OX = XK ; 

\d) ZAYZ = ZB= ZAOZ; 
(e) As AYZ, BZX, CXY, ABC are similar; 
{/) AX, BY, CZ bisect the As of the pedal A XYZ; 
(g) of the four points A, B, C, O, each is the orthocentre 
of the A of which the other three points are the vertices; 



EXERCISES 7 

(h) if a A LMN be formed by drawing through A, B, C 
lines MN, NL, LM II BC, CA, AB respectively, O is the 
circumscribed centre of A LMN; 

(i) if S be the centre of the circumscribed circle of A 
ABC, AS, BS, CS are respectively J_ YZ, ZX, XY the sides 
of the orthocentric A. 

7. In A ABC, the inscribed circle touches BC, CA, AB 
at D, E, F respectively, and s = the semi-perimeter. Show 
that AF = s - a. 

If a = 43, b = 31, c = 26, show that AF . BD . CE = 3192. 
X8. The st. lines joining the vertices of a A to the points 
of contact of the opposite sides with the inscribed circle 
are concurrent. 

i). In A ABC, an escribed circle touches BC at D and 
AB, AC, produced at F, E respectively. Show that FB = 
s - c. 

If a = 40, 6 = 30, c = 50, show that AF . BD.CE= 18000. 
*>10. The st. lines joining the vertices of a A to the 
points of contact of the opposite sides with any one of the 
escribed circles are concurrent. 

11. O is a point within the A ABC and AO, BO, CO 
produced cut BC, CA, AB at D, E, F respectively. The 
circle through D, E, F cuts BC, CA, AB again at P, Q, R. 
Show that AP, BQ, CR are concurrent. 

(*) 

<*12. The bisectors of L s B, C of A ABC cut CA, AB at 
E, F respectively. FE, BC produced meet at D. Prove 
that AD bisects the exterior Z. at A. 

*13. The points where the bisectors of the exterior /-s at 
A, B, C of A ABC meet BC, CA, AB respectively are 
collinear. 



8 



SYNTHETIC GEOMETRY 



14. AB, CD, EF are three )| st. lines. AC, BD meet at 
N; CE, DF at L; EA, FB at M. Prove that L, M, N are 
collinear. 

15. (a) If two As are so situated that the st. lines 
joining their vertices in pairs are concurrent, the inter- 
sections of pairs of corresponding sides are collinear. — 
Desargues' Theorem. 

(Note. — ABC, abc the two As; Aa, Bb, Cc meeting at OJ 
BC, ic at L; CA, ca at M ; AB, ab at N. Using the As 
OBC, OCA, OAB and the respective transversals bcL, acM, 
a&N prove that AN . BL . CM = NB . LC . MA.) 

(b) State and prove the converse of (a). 

(Note.— BC, JcmeetatL; 
CA, ca at M ; AB, ab at N 
and L, M, N are collinear. 
Produce Act, Bb to meet at 
O. To show that Cc passes 
through O. A«M, B&L 
are As having AB, ab, ML 
concurrent at N ; corres- 
ponding sides aM, bLmeet 
at c; MA, LB at C; Aa, 
Bb at O. .*., by (a), c, 
C, O are collinear, i.e., Aa, 
N Bb, Cc are concurrent.) 

Fig. 11. 

(Historical Notb :— Girard Desargues (1593-1662) was an architect and engineer of 
Lyons, France.) 

16. The inscribed circle of A ABC touches the sides 
BC, CA, AB at D, E, F respectively; EF, FD, DE, produced 
meet BC, CA, AB respectively at L, M, N. Show that 
L, M, N are collinear. 

(Note.— Use Ex. 8 and Ex. 15 (a).) 




EXERCISES 9 

17. Tangents to the circumcircle at A, B, C, meet BC, 
CA, AB respectively in collinear points. 

(Note.— Use Ex. 8 and Ex. 15 (a).) 

18. Given the base and vertical Z. of a A, find the locus 
of its orthocentre. 

19. If the base BC and vertical L A of a A ABC be 
given, and the base be trisected at D, E, the locus of the 
centroid is an arc containing an L equal to L A, and 
having DE as its chord. 

20. ABC is a A, XYZ its pedal A. Show that the 
respective intersections of BC, CA, AB with YZ, ZX, XY 
are collinear. 

21. "Where is the orthocentre of a rt.-Z-d A 1 ? 

22. If one escribed circle of A ABC touch AC at F and 
BA produced at G, and another escribed circle touch AB 
at H and CA produced at K, FH, KG produced cut BC 
produced in points equidistant from the middle point of BC. 

(Note. — Use § 1, taking the two transversals FH, KG of 
A ABC and multiplying the results.) 

23. If O is the orthocentre, S the circumcentre and I 
the centre of the inscribed circle of A ABC, prove that 
IA bisects L OAS. 



10 



SYNTHETIC GEOMETRY 




8. The distance from each vertex of a triangle 
to the orthocentre is twice the perpendicular from 
the circumcentre to the side opposite that vertex. 

O is the orthocentre, S the 
circumcentre of A ABC; SD 
is ± BC. 

To show that AO = twice 
SD. 

Draw the diameter CSE, 
join BE, EA. 

Ls EBC, EAC being Ls in 
fig. 12. semicircles are rt. Ls. 

;. EB !| AO and EA II BY. 
.*. EAOB is a || gm, and EB = AO. 
But V S, D are middle points of EC, BC, 

.*. EB = twice SD, 
and /. AO = twice SD. 

9. ABC is a A having AX _L 
BC, AD a median, O the ortho- 
centre, and S the circumscribed 
centre. Show that OS cuts AD 
at the centroid G. (Use § 8.) 
Show also that G is a point 
of trisection in SO. 

A general enunciation of these 
results maybe given as follows: — 

The Orthocentre, Centroid, and the centre of the 
Circumscribed Circle of a A are in the same st 
line, and the Centroid is a point of trisection in the 
St. line joining the other two. 




the nine-point circle 
The Nine-Point Circle 



11 



10. The three middle points of the sides of a 
triangle, the three projections of the vertices on the 
opposite sides, and the three middle points of the 
straight lines joining the vertices to the orthocentre 
are all concyclic. 




Let ABC be a A, AX, BY, CZ the J_s from A, B, C 
to BC, CA, AB respectivel}', O the orthocentre, L, M, N 
the middle points of AO, BO, CO respectively, D, E, F 
the middle points of BC, CA, AB respectively. 

It is required to show that the nine points, X, Y, Z, 
L, M, N, D, E, F are concyclic. 

Find S, the clrcumcentre. Draw SO, SD, SA. Biseci 
SO at K. Draw KL. 

V K, L are the middle points of SO, OA ; 

.'. KL = half of SA; 
and .*. the circle described with centre K and radius 
equal to half that of the circumcircle passes through L. 



12 SYNTHETIC GEOMETRY 

Similarly this circle passes through M and N. 

Draw DK. 

f SD= LO, (§8) 

In AS SKD, OKL^ SK= KO, 

(zDSK=ZKOL, (SDULO); 

.'. DK = KLand ZSKD= ZOKL. 

DK = KL; the circle with centre K and radius 
KL passes through D. 

Similarly this circle passes through E and F. 

ZSKD= ZOKL, and SKO is a st. line; 

LKD is a st. line 

K is the middle point of the hypotenuse of the 
rt.-Zd A LDK; 

the circle with centre K and radius KL passes 
through X. 

Similarly this circle passes through Y and Z. 

the nine points L, M, N, D, E, F, X, Y, Z are 

concyclic. 

Cor. 1: — The centre of the N.-P. circle is the middle 
point of the line joining the circumcentre to the ortho- 
centre. 

Cor. 2: — The diameter of the N.-P. circle is equal to 
the radius of the circumcircle. 






simpson s line 
Simpson's Line 



13 




11. If any point is taken on the circumference of 
the circumscribed circle of a triangle, the projections 
of this point on the three sides of the triangle are 
collinear. 

Let P be any point on the 
circle ABC, X, Y, Z, the pro- 
jections of P on BC, CA, AB 
respectively. 

It is required to show that 
X, Y, Z are in the same st. 
line. 

Join ZY, YX, PC, PA. 
Z PYC = L PXC, /. P, Y, 
X, C, are concyclic, and L 
XYC = L XPC. 

Z AZP+ Z AYP = 2 rt. Zs, 
and Z AYZ = Z APZ. 

APCB is a cyclic quadrilateral, 

.-. Z APC + Z B = 2 rt. 
la quadrilateral BZPX, 

Zs BZP, BXP arert. Zs, 
.-. Z ZPX+ z B = 2 rt. Zs. 
Hence Z ZPX = Z APC, and as the 
common to these Zs, 

L APZ = L XPC. 
.'. Z AYZ = L XYC, and 
L XYC + Z CYZ = L AYZ + CYZ = 2 rt. As \ 
:. XY and YZ are in the same st. line. 

(Historical Notb.— Robert Simpson (1687-1768) was Professor of Mathematics in 
the University of Glasgow.) 



/. A, Z, P, Y are concyclic, 



Zs. 



part APX is 



14 SYNTHETIC GEOMETRY 

12.— Exercises 
(a) 

1. P is the orthocentre of A DEF, and the ||gm EPFG 
is completed. Show that DG is a diameter of the circle 
circumscribing DEF. 

2. ABC is a A ; L, M, N the centres of its escribed 
circles. Show that the circle circumscribed about ABC is 
the N.-P. circle of A LMN. 

3. In A ABC, I is the centre of the inscribed circle, 
L, M, N the centres of the escribed circles. Prove that the 
circumcircle of A ABC bisects IL and LM. 

4. O is the orthocentre of A ABC. Prove that As OBC, 
ABC have the same N.-P. circle. 

5. Given the base and vertical Z of a A, show that the 
locus of the centre of its N.-P. circle is a circle having its 
centre at the middle point of the base. 

(Note. — Using Cor. 2 of § 10 show that the distance of 
K from the fixed point D is constant.) 

6. If the projections of a point on the sides of a A are 
collinear, the point is on the circumcircle of the A. 

7. The three circles which go through two vertices of a 
A and its orthocentre are each equal to the circle circum- 
scribed about the A. 

8. The ± from the middle point of a side of a A on 
the opposite side of the pedal A bisects that side. 

9. Construct a A given a vertex, the circumcircle and 
the orthocentre. 

(Note. — Describe the circumcircle and produce AO to 
cut it in K, where A is the given vertex and O is the ortho- 
centre. Draw the rt. bisector of OK meeting the circle 
in B, C. See § 7, Ex. 6 (c).) 



EXERCISES 15 

10. DEF is a A and O is its orthocentre. About DOF 
a circle is described and EO is produced to meet the 
cirumference at P. Show that DF bisects EP. 

11. I is the centre of the inscribed circle of A ABC, and 
Al, Bl, CI are produced to meet the circumcircle at L, M, N. 
Prove that I is the orthocentre of A LMN. 

12. I is the centre of the inscribed circle of A ABC, and 
the circumcircle of A IBC cuts AB at D. Prove that 
AD = AC. 

13. In A ABC, the _Ls from A, B to the opposites sides 
meet the circumcircle at D, E. Show that arc CD = arc CE. 

14. XYZ is the pedal A of A ABC. Prove that A, B, C 
are the centres of the escribed circles of A XYZ. 

(*) 

15. X, Y, Z are the projections of A, B, C on BC, CA, 
AB. Prove that 

(a) YZ . ZX = AZ . ZB ; 

(b) YZ . ZX . XY = AZ . BX . CY. 

16. Given the base and vertical L of a A, to find the 
loci of the centres of the escribed circles. Let BC be the 
base and BDC a segment of a circle containing L BDC = 
the given vertical L. Draw the rt. bisector of BC cutting 
the circle BDC at D, E. Prove the loci are arcs on the 
chord BC and having their centres at D, E. 

17. Construct a A having given the base, the vertical Z 
and the radius of an escribed circle. (Two cases.) 

(Note.— Construct the loci as in Ex. 16. In one case, on 
the arc with centre E, find the point l a such that its dis- 
tance from BC equals the given radius. Draw l x E and 
produce to cut the arc BDC at A. ABC is the required A. 



16 



SYNTHETIC GEOMETRY 



In the other case on the arc with centre D find the 
point l 2 such that its distance from BC equals the given 
radius. Draw l 2 D cutting the arc BDC at A'. A'BC is 
the required A.) 

18. O is the orthocentre of A ABC, and D, E, F are the 
circumcentres of As BOC, COA, AOB. Show that 

(a) A DEF = A ABC; 

(b) The orthocentre of each of the As ABC, DEF is 

the circuincentre of the other ; 

(c) The two As have the same N.-P. circle. 

(Note. — The rt. bisectors of AO, BO, CO form the 
sides of A DEF. Let L be the middle point of AO. J_s 
from D, E on BC, CA respectively meet at S the circum- 
centre of A ABC. DS bisects BC at G. Prove L LEO = 
L OCA = L GCS and comparing As OLE, GSC, using §8, 
show LE = GC. Similarly LF = BG, so that FE = BC, etc.) 

19. Find a point such that its projections on the four 
sides of a given quadrilateral are collinear. 




,'P^ 



(Note. — ABCD the given quadrilateral; produce the 
opposite sides to meet at E, F. Describe circles about 
As EBC, FCD meeting again at P. P is the required point. 



EXERCISES 



17 



20. In the A ABC, the _L from A to BC is produced to 
cut the circumcircle at P. Prove that the Simpson's Line 
of P is || to the tangent to the circumcircle at A. 

21. P is any point on the circumcircle of A ABC. The 
J_s from P to the sides of the A meet the circle at D, E, F. 
Prove that A DEF = A ABC. 

22. P is any point on the circumcircle of a A ABC of 
which O is the orthocentre and X the projection of A on 
BC ; AX produced cuts the circumcircle at D and PD cuts 
BC at E. Prove that the Simpson's Line of P bisects PE, 
is || OE, and bisects OP. 

(Note. — ML cuts PE at F. 
Draw PC. 

Z FPL = Z FDA = Z PCA = 
Z PLF; 

/. FL = FP. Then F is the 
middle point of the hypotenuse 
of the rt.-Zd A PLE. 

.*. Simpson's Line bisects PE. 

V ZFLP= ZXDE; .*. Z FLE 
= Z XED. But Z OEX= L XED 
(See § 7, Ex. 6 (c). 

.'. L OEX = Z FLE, and LM [| OE. 

In A POE, LM || OE and bisects PE.: .'. LM bisects PO.) 

Give a general statement of the last of the three results 
in Ex. 22. 





18 synthetic geometry 

Areas of Rectangles 
13. If from the vertex of a triangle a straight line 
is drawn perpendicular to the base, the rectangle 
contained by the sides of the triangle is equal to 
the rectangle contained by the perpendicular and 
the diameter of the circumcircle of the triangle. 

AX ± BC and AD is a diame- 
ter of the circumcircle of A ABC. 
To prove that 

rect. AB . AC = rect. AX. AD. 
Join DC. 

V Z AXB = Z ACD, 

and Z ABX = Z ADC. 

,\ A AXB HI AACD. 

AB AX 

no. ib. • = . 

AD AC 

.-. rect. AB . AC = rect. AX . AD. 

H. If the vertical angle of a triangle is bisected 
by a straight line which also cuts the base, the 
rectangle contained by the sides of the triangle is 
equal to the rectangle contained by the segments 
of the base together with 
the square on the straight 
line which bisects the angle. 

ABC is a A and AD the 
bisector of L A. 

It is required to show b 1 
that the rect. AB . AC = rect. 
BD . DC + AD 2 . 

Circumscribe a circle about 
the A ABC. Produce AD to 
cut the circumference at E. Join EC. 




AREAS OF RECTANGLES 



19 



111 As BAD, EAC Z BAD = Z EAC, Z ABD = Z AEC, 
.*. Z ADB = Z ACE and the As are similar; 
i BA EA 

heDCe AD = AC' 

and .\ BA . AC = AD . EA. 
But AD . EA = AD (AD + DE) 

= AD 2 + AD . DE 
= AD 2 + BD. DC. 
.*. rect. BA . AC = rect. BD . DC + AD'. 
15. Ptolemy's Theorem: — The rectangle contained 
by the diagonals of a quadrilateral inscribed in a 
circle is equal to the sum of the rectangles contained 
by its opposite sides. 

This theorem is a particular case 
of that of % 16. 

A BCD is a quadrilateral in- 
scribed in a circle. 
To prove that 

AC . BD = AB . CD + BC . AD. 
Make Z BAE = Z CAD, and 
produce AE to cut BD at E. 

A ABE U| A ACD, 

AB _ BE 

AC CD' 

AB.CD = AC. BE. 

A ADE III A ABC, 

DE 

BC* 




FiO. 20. 



AD 
AC 



AD.BC = AC. DE. 



AB.CD + AD. BC 



= AC. BE + AC . DE 
= AC (BE + ED) 
= AC . BD. 



(Historicai, Note.— Ptolemy, a native of Egypt, flourished in Alexandria in 
A.l>.) 



SYNTHETIC GEOMETRY 



16. The sum of the rectangles contained by the 
opposite sides of a quadrilateral is not less than the 
rectangle contained by the diagonals. 

ABCD is a quadrilateral, 
AC, BD its diagonals. 

Required to show that 
AB . DC + AD . BC is not 
less than AC . BD. 

Make ^ BAE = z CAD 
and Z ADE = L ACB. 
Join EB. 
As BAC, EAD are similar, 



L CAD, 

.*. As BAE, CAD are also similar. 
From the similar As BAE, CAD 

and .*. AB . CD = AC . BE. 




AB 
BE 



AC 
CD 



From the similar As BAC, 
BC ED 
AC ~ AD' 



id .*. BC 



EAD, 
AD = 



AC . ED. 



Consequently AB . CD + BC . AD = AC (BE 4- ED) 
but BE + ED is not < BD ; 
.*. AB , CD 4 BC . AD is not < AC . BD, 



EXERCISES 



21 



J&^l 



17.— Exercises 

(a) 



1. If the exterior vertical Z A of A ABC be bisected by 
a line -which cuts BC produced at D, rect. AB . AC = rect. 
BD . CD - AD 2 . 

(Note. — Draw the circle 

ACB. Produce DA to cut 

the circumference at E. Draw 

EC. ProveABAD ;;ZaEAC, 

BA EA _. 
and that ,\ -,-=: = Tq* lhen 

BA . AC = AD . EA = AD 
(ED - AD) = AD . ED - AD 2 
= BD. CD - AD 2 .) 

2. Draw A ABC having a — 81 mm., b = 60 mm., c = 30 
mm. Bisect the interior and exterior Zs at A and produce 
the bisectors to meet BC and BC produced at D and E. 
Measure AD, AE ; and check your results by calculation. 

3. If the internal and external bisectors of Z A of A ABC 
meet" 3C at L, M respectively, prove 

LM 2 = BM . MC - BL. LC. 

4. If R is the radius of the circumcircle and A the area 
of A ABC, prove that 

abc 




R = 



4 A 



If the sides of a A are 39, 42, 45, show that R = 24f. 
5. P is any point on the circumcircle of an equilateral 



A ABC. Show that, of the three distances PA, 
one is the sum of the other two. 



PB, PC, 



(») 

6. From any point P on a circle ±s are drawn to the 
four sides and to the diagonals of an inscribed quadrilateral. 
Prove that the rect. contained by the ±s on either pair of 



82 



SYNTHETIC GEOMETRY 



opposite sides is equal to the rect. contained by the J_s on 
the diagonals. 

7. "With given base and vertical Z construct a A having 
the rect. contained by its sides equal to the square on a 
given st. line. 

8., A, B, C, D are given points on a circle. Find a point 
P on the circle such that PA . PC = PB . PD. 

9. AB is the chord of contact of tangents drawn from a 
point P to a circle. PCD cuts the circle at C, D. Prove 
that AB . CD = 2 AC . BD. 

10. I is the centre of the inscribed circle of A ABC. Al 
produced meets the circuracircle at K. Prove Al. IK = 2Rr. 

(Note. — Draw the diameter KE of 
circle ABC and the radius IN of the 
inscribed circle. Draw BE, BK, Bl. 

Show that A BEK III A NAI, and 

rt. * . Al KE - 

that .. W = Kg. 

or, Al . KB = 2Rr. 
Show that KB = Kl, and that .«. 
Al . IK = 2Rr.) 

Hence, using Ex. 6, Page 256, 
O.H.S. Geometry, show that, if S be the circumcentre of 
A ABC, SI 2 = R 2 - 2Rr. 

11. I x is the centre of the escribed circle opposite to A 
in A ABC. Ah cuts the circumcircle ABC at K. Prove 
Al 2 . I X K = 2Rr x . 

Hence show that, if S be the circumcentre of A ABC, 
Sli 3 = R 2 + 2R/v 




RADICAL AXIS 



23 



Radical Axis 

18. The locus of the points from which tangents 
drawn to two circles are equal to each other is called 
the radical axis of the two circles. 

19. If two circles cut each other, their common 
chord produced is the radical axis. 

[Proof left for the pupil] 








20. The locus of a point P 
such that the difference of the 
squares of its distances from 
two fixed points A, B is 
constant is a st. line perpen- 
dicular to AB. FlQ 25 

From P draw PM j_ AB. Let AB = a, AM 
PA 2 - PB 2 = k, where a and k are constants. 
AM 2 4- MP 2 = PA 2 
MB 2 +MP 2 = PB 2 
.*. AM 2 - MB 2 = PA 2 - PB 2 = h. 
or x 2 - (a - x) 2 = k. 
a 2 +k 



and 



. 






and x = 



2a 



24 



SYNTHETIC GEOMETRY 



Hence AM is constant and M is a fixed point. 

.*. the locus of P is a st. line 1 AB drawn through 
the fixed point M. 

21. A, B are the centres of two circles of radii 
R, r respectively. 

To prove that the radical axis of the circles is a 
st. line _L AB and cutting it at a point M such that 

AM 2 - MB 2 = R 2 - r\ 




Let P be any point on the radical axis, and draw 
PM ± AB. 

Draw the tangents PC, PD to the circles, and join 
PA, PB, AC, BD. 

PA 2 = PC 2 + R 2 , 

PB 2 = PD 2 + r 2 , 
and, since P is on the radical axis, PC = PD; 
.-. PA 2 - PB 2 = R 2 - r 2 , a constant. 
.•., by § 20, the locus of P is a st. line _L AB. 
Also PA 2 = AM 2 +PM 2 ; 
PB 2 = MB 2 +PM 2 l 
.♦. PA 2 - PB 2 = AM 2 - MB 2 ; 
and .'. the radical axis cuts AB at the fixed point M, 
such that AM 2 - MB 2 = R 2 - r 2 . 



EADICAL AXIS 



25 



22. To draw the radical axis of two non-inter- 
secting circles. 




Let A, B be the centres of the circles. 

Describe a circle with centre O cutting the given 
circles at C, D and E, F. Draw CD, EF and produce 
them to meet at P. Draw PM j_ AB. Draw tangents 
PG, PH to the circles. 

Show that PM is the required radical axis. 






SYNTHETIC GEOMETRY 
Second Method 




Let A, B bo the centres of the two circles. 

Join AB. Through B draw BC _L AB cutting the 
circle with centre B at C, and cut off BD equal to 
the radius of the other circle. 

With centre A and radius AD describe an arc, and 
with centre B and radius AC describe another arc 
cutting the first at E. Draw EM J_ AB. 

BM 2 = BE 2 - EM 2 = AB 2 + BC 2 - EM 2 . 

MA 2 = AE 2 - EM 2 = AB 2 + BD 2 - EM 2 . 

.*. BM 2 - MA 2 = BC 2 - BD 2 . 

•'• > by § 21, EM is the radical axis. 



EXERCISES 27 

23.— Exercises 

1. Draw two circles, radii 1 inch and 2 inches, with 
their centres 4 inches apart. Find a point whose tangents 
to the two circles are each 1| inches in length. 

2. The radical axis of two circles bisects their common 
tangents. 

3. Find the radical axis of two circles which touch each 
other, internally or externally. 

i. Prove that the radical axes of any three circles taken 
two and two together meet in a point. 

Note. — This point is called the radical centre of the 
three circles. 

5. O is a fixed point outside a given circle. P is any 
point such that the tangent from P to the given circle = 
PO. Show that the locus of P is a st. line J_ to the line 
joining O to the centre of the circle. 

<& C is a point on the circumference of a circle with 
centre A. Join AC and draw CB J_ CA. "With centre B 
and radius BC describe a circle. 

(a) The tangents to the circles at a common point 

are -L to each other. 

(b) The square on the line joining the centres of the 

circles equals the sum of the squares on their radii. 

Definition. — Circles which cut each other so that the Js 
tangents at a common point are at right angles to each 
other are said to be orthogonal. 

7. If O be the orthocentre of A ABC, the circles described 
on AB and CO as diameters are orthogonal. 

8. Tf circles are described on the three sides of a A as 
diameters, their radical centre is the orthocentre of the A. 



28 



SYNTHETIC GEOMETRY 



9. Through two given points A and B draw any number 
of circles. What is the locus of their centres? Show that 
any two of this system of circles have the same st. line for 
radical axis. 

Definition — A system of circles that have the same 
radical axis are said to be coaxial. 

10. To draw a system of circles coaxial with two given 
non-intersecting circles. 

Let A, B be the centres of the given circles. Draw their 
radical axis PO cutting AB at O. From O draw a tangent 
OE to either circle. With centre O and radius OE describe 
a circle cutting AB at C, D. On the circle CDE take any 
point F and at F draw a tangent to the circle CED cutting 
AB at G. With centre G and radius GF describe a circle. 
Prove that this circle is coaxial with the given circles. 

By taking different positions of F on the circle CED any 
number of circles may be drawn coaxial with the given 
oircles. 




Definition. — No circle of the coaxial system has its centre 
between C and D, and consequently these points are called 
the limiting points of the system. 



EXERCISES 29 

11. In Fig. 29, show that :— 

(a) The circle CED cuts each circle of the coaxial 

system orthogonally; 

(b) Any circle with centre in PO and passing through 

C, D cuts any circle of the coaxial system ortho- 
gonally. 

(*) 

12. If from any point P tangents be drawn to two circles, 
the difference between their squares equals twice the rect- 
angle contained by the _L from P on the radical axis of 
the two circles and the distance between their centres. 

13. The difference of the squares of the tangents drawn 
from a point to two fixed circles is constant. Show that 
the locus of the point is a st. line ± to the line of centres 
of the circles. 

14. The tangent drawn from a limiting point to any 
circle of a coaxial system is bisected by the radical axis. 

15. Show that the locus of the centre of a circle, the 
tangents to which from two given points are respectively 
equal to two given st. lines, is the radical axis of the circles 
having the given points as centres and radii respectively 
equal to the two given st. lines. 

16. With a given radius describe a circle to cut two 
given circles orthogonally. 

17. XYZ is the pedal A of A ABC; YZ, BC meet in L; 
ZX, CA meet in M ; XY, AB meet in N. Show that L, M, N, 
are on the radical axis of the circumscribed and N.-P. 
circles of A ABC. 

(Note. — As MAZ, MXC are easily shown to be similar.) 

18. Describe a circle to cut three given circles orthogonally. 
(Note.— Use Ex. 4 and Ex. 11 (&).) 



CHAPTER II 
Medial Section 

,53". When a straight line is divided into two parts 

such that the square on one part is equal to the 

rectangle contained by the 

"**"v**< v given straight line and the 

\ \ other part, the straight line 

is said to be divided in 

j '• \ medial section. 

S F ..-•'" (25^ To divide a given 

„..-**" straight line internally in 

i ..-'' medial section. 

D;'" Let AB be the given st. 

: line. 

Draw AC j_ AB and = AB. 
£■ Bisect AC at D. With centre 

Fl °- 30 - D and radius DB describe an 

arc cutting CA produced at E. With centre A and 
radius AE describe an arc cutting AB at F. 
Then AB is divided in medial section at F. 
DA 2 +AB 2 = DB 2 
= DP 

= DA 2 + 2 DA. AE + AE 2 
= DA 2 + AB. AF + AF 2 ; 
V 2 DA = AC = AB and AE = AF. 
.*. AF 2 = AB 2 - AB . AF. 
= AB (AB - AF) 

= AB . BF. 

30 



MEDIAL SECTION 



31 



26. To divide a given straight line externally in 
medial section. 

Let AB be the given st. line. 



A 
D 
C 

E ,. 



^'B\ 



Draw AC _L AB and = AB. Bisect AC at D. With 
centre D and radius DB describe an arc cutting AC 
produced through C at E. With centre A and radius 
AE describe an arc cutting BA produced through 
A at F. 

Then AB is divided externally at F such that 
AF 2 = AB . BF. 

DA 2 + AB 2 = DB 2 

= DE 2 = (AE - AD) 2 

= AE 2 - 2 AE. AD + AD 2 

= AF 2 - AF. AB + AD 2 

AF 2 = AB 2 + AF. AB 

= AB (AB + AF) 

= AB . BF. 



32 



SYNTHETIC GEOMETRY 



27.— Exercises 

1. If a st. line AB be divided at F so that AP= AB.BF, 
show that AB : AF = AF : BF. 

Give a general statement of this result. 

2. AB is divided internally at F such that AF 3 = AB . BF. 
Show that AF > FB. 

3. AB is divided in medial section at 
F. On AB, AF squares ABCD, AFEG 
are described as in Fig. 32. EF is pro- 
duced to meet DC in H. Show that 
the rectangle DE = the square DB. 

4. In Fig. 32 join GB, DF, pro- 
duce DF to meet GB, and show that 
DF _[_ GB. 

5. A given st. line AB is to be 
divided in medial section. Let F be 

Fl0 3 2 the point of section, a the length of 

AB, x the length of AF. 

Then, by the definition of medial section, x 2 = a (a — x) 

or, x- -f ax - a 2 = 0. 

c , . ,, . n . . - a ± a i/5 
feolving this quadratic equation, a; = • 



G 


E 




F 


A 




D 


H 



Show that the construction in § 25 is sugg 
a -\- a V5 



a - a |/5 



ted by the root 
and the construction in § 26 by the root 



6. Divide a st. line 4 inches in length in medial section. 
Measure the length of each part, and test the results by 
calculation. 

7. The difference of the squares on the parts of a st. line 
divided in medial section equals the rectangle contained by 
the parts. 



MEDIAL SECTION 33 

'8. If AB be divided at C so that AC 2 = AB . BO, show 
that AB 2 + BC 2 = 3 AC 2 . 

'9. If the sides of a rt.- Zd A are in continued proportion, 
the J- from the rt. Z divides the hypotenuse in medial 
section. 

10. Describe a rt.-Zd A "whose sides are in geometrical 
progression. 



28. To describe an isosceles triangle having each 
of the angles at the base double the vertical angle. 

Draw a st. line AB and divide it at H so that 
AH' = AB.BH. (§25.) 

Describe arcs with centres A, B and radii AB, AH 
respectively, and let them cut at C.^^ 

Join AC, BC. /A 

ABC is the required A. 

Join HC. 

Z B is common to the As ABC, CBH, 



and since 




AB AH . AB BC 
AH BH' •'• BC BH' 


B C 

FlQ. 33. 


.*. these As are similar and z BCH = 


= Z A. 


. . AC A B AH 
A g ain BC = AH = HB' 





CH bisects Z ACB. 
Z ACB = twice Z BCH = twice Z A; 
and also Z ABC = twice Z A. 



34 



SYNTHETIC GEOMETRY 



L'9.— Exercises 

(«) 

1. Express the Zs of the A ABC (Fig. 33) ia degrees. 

2. Construct Zs of 36°. 18°, 9°, 6°, 3°. $1° 

* 3. Show that AHC (Fig. 33) is an isosceles A having the 
vertical Z three times each of the base Zs. 

i. Show that each Z of a regular pentagon is 108° ; and 
that, if a regular pentagon is inscribed in a circle, each 
side subtends at the centre an Z of 72°. 

* 5. In a given circle CAB draw 
any radius OA. Divide AO at H 
so that OH 2 = AO . AH. Place the 
chord AB = HO. 

Join BH and produce BH to cut 
the circumference at C. Join AC. 

Show that AB is a side of a 
regular decagon inscribed in the 
circle ; and that AC is a side of 
a regular pentagon inscribed in 
the circle. 

J6J In a given circle inscribe a regular pentagon. At the 
angular points of the pentagon draw tangents meeting at 
A, B, C, D, E. Show that ABODE is a regular pentagon 
circumscribed about the circle. 

/T. Show that each diagonal of a regular pentagon is || 
to one of its sides. 

8. Draw a regular pentagon on a given st. line. 

9. ABCDE is a regular pentagon. Show that AD, BD 
trisect Z CDE. 




EXERCISES 35 

10. ABCDE is a regular pentagon. Show that AC, BD, 
divide each other in medial section. 

11. Construct a regular 5-pointed star. What is the 
measure of the Z at each vertex? 

0) 

12. Show that the side of a regular decagon inscribed in 

f 

a circle of radius r is — (]/ 5 - 1). 

13. Show that the side of a regular pentagon inscribed in 
a circle of radius r is — -»/ 10 - 2y 5. 

14. The square on a side of a regular pentagon inscribed 
in a circle equals the sum of the squares on a side of the 
regular inscribed decagon and on the radius of the circle. 

15. In a circle of radius 2 inches inscribe a regular decagon 
by the method of Ex. 5. Measure a side of the decagon and 
check your result by calculation. 

16. In a circle of radius 3 inches inscribe a regular pentagon 
by the method of Ex. 5. Measure a side of the pentagon and 
check your result by calculation. 

17. In a given circle draw two radii OA. OB at rt. Zs to 
each other. Bisect OB at C. ^"""~^"**"»>^ 

Join AC, and cut off CD = CO. yf ^v 

Show that AD is equal to a / \ 

side of a regular decagon in- / \ 

scribed in the circle. [~""" r "P 

The regular inscribed pentagon \ ;\ / J 

may be drawn by joining alter- \ f / / 

nate points obtained by placing N. ;/ yf 
successive chords each equal to ^ 

AD. Fio. 35. 

18. On a st. line 2 inches in length describe a regular 
pentagon. Measure a diagonal of the pentagon and check 
your result by calculation. 



db SYNTHETIC GEOMETRY 

19. If the circumference of a circle be divided into n 
equal arcs, 

(a) The points of division are the vertices of a regular 
polygon of n sides inscribed in the circle; 

(6) If tangents be drawn to the circle at these points, 
these tangents are the sides of a regular polygon 
of n sides circumscribed about the circle. 

20. Show that the difference between the squares on a 
diagonal and on a side of a regular pentagon is equal to 
the rectangle contained by them. 



tn/3f>*n/°C 



MISCELLANEOUS THEOREMS 37 



Miscellaneous Theorems 

30. ABC is a triangle and P is a point in BC 

such that !!£ = —. It is required to show that 
PC ni 

mAB 2 + nAC* = (m + n) AP 2 4- mBP 2 + nCP\ 
Draw AX jl BC. 

From A ABP, /i\V 

AB 2 = AP 2 + BP 2 - 2 BP.PX. /] VV 

From A APC, / j \ \ 

AC 2 = AP 2 +CP 2 +2 CP, PX. B X p Q 

Fig. 36. 

Multiplying both sides of 
the first of these equations by m, both sides of the 
second by n, adding the results and using the condition 
??i BP = tiPC, we obtain 

mAB 2 + nAC 1 = (m + n) AP 2 + mBP 2 + ?iCP 2 . 
What does the result in § 30 become when m = n ? 

In a A ABC, a = 77 mm, 6 = 90 mm and c = 123 
mm. Find the distances from C to the points of 
trisection of AB. 

(31. In a right-angled triangle a rectilineal figure 
described on the hypotenuse equals the sum of the 
similar and similarly described figures on the other 
two sides. 

ABC is a A rt.-Zd at C and having the similar and 
similarly described figures X, Y, Z on the sides. 

It is required to show that X + Y = Z. 



38 



SYNTHETIC GEOMETRY 



Similar figures are to each other as the squares on 
corresponding sides. 






Y 

•• Z 


AC 2 
~ AB'' 




and =- 


BC 2 
~ AB 2 ' 




X + Y 
Z 


AC 2 +BC 2 
AB 2 


But 


AC 2 + BC 2 


= AB 2 . 




.'. X + Y 


= Z. 



Prove this theorem by drawing a J_ from C to AB 
and using the theorem : — If three st. lines are in 
continued proportion, as the first is to the third so is 
any polygon on the first to the similar and similarly 
described polygon on the second. 



SIMILAR AND SIMILARLY SITUATED POLYGONS 39 



Similar and Similarly Situated Polygons 

f Similar polygons are said to be similarly situ- 
when their corresponding sides are parallel and 
drawn in the same direction from the corresponding 
vertices. 

33. If two similar triangles have their corresponding 
sides parallel, the st. lines joining corresponding vertices 
are concurrent. 

Let ABC, DEF be two similar As having the sides 
BC, CA, AB respectively || to the corresponding sides 
EF, FD, DE. ~ oc.or 

c r 





Prove AD, BE, CF concurrent. 

('34.; When two similar polygons are so situated that 
their corresponding sides are parallel but drawn in 
opposite directions from the corresponding vertices, 
they are said to be oppositely situated. 

In Fig. 39, the similar As ABC, DEF are oppositely 
situated. 



40 



SYNTHETIC GEOMETRY 



35. If two similar polygons have the sides of one 
respectively parallel to the corresponding - sides of 
the other, the straight lines joining corresponding 
vertices are concurrent. 





Let ABCDE, abcde be two similar polygons, similarly 
situated in Fig. 40, oppositely situated in Fig. 41. 

Join Aa, B6 and let the joining lines meet at O. 
Join CO cutting he at x. 



From similar As, 






ah 
AB 


Oh 
OB 


hx 
= BC 



EXERCISES 41 

But, by hypothesis, 

ab be 
AB = BC* 
bx = be, and 
OC passes through c. 
Similarly it may be shown that the st. lines joining 
the remaining pairs of corresponding vertices pass 
through O. 

36.— Exercises 

(a) 
X. Inscribe a square in a given A. Show that there are 
three solutions. 

(Note. — In A ABC on BC externally describe the square 
BDEC. Draw AD, AE cutting BC at F, G respectively. 
Draw FH, GK _L BC meeting BA, CA at H, K respectively. 
Draw HK. Prove that HFGK is a square.) 

2. In a given A inscribe a rectangle similar to a given 
rectangle. Show that there are six solutions. 

3. In a given semicircle inscribe a square. 

"i Ina given semicircle inscribe a rectangle having its 
sides in a given ratio. 

"^ In a given A inscribe a A having its sides || to three 
given st. lines. 

(Note. — From any point D in the side BC of the 
given A ABC draw DE || to one of the given st. lines and 
meeting AC at E. Draw DF, EF respectively || to the other 
given lines. Draw CF cutting AB at G. Draw GH || FD 
and GK || FE, meeting BC, AC respectively at H, K. Draw 
HK. Show that GHK is the required A.) 

(*) 

6. The base of a square lies on one given st. line and one 
of its upper vertices lies on another given st. line. Show that 
the locus of the other upper vertex is a fixed st. line passing 
through the point of intersection of the two given lines. 



42 SYNTHETIC GEOMETRY 

7. In Figures 40 or 41 P is any point in AB, Q is any 
point in CD and p, q are the corresponding points in ab, 
cd respectively. Prove that PQ || pq. 

8. A ABC HI A «& c » with their corresponding parts in the 
same circular order, but their corresponding sides are not ||. 
BC, be meet at P. The circles BPb, CPc meet again at O. 
Prove that the A abc may be rotated about O to a position 
where it is similarly situated to A ABC. 

(Note.— Prove L BOb = Z COc - L AOa.) 




Fig. 42. 

9. Show that 4R = r x -\-r 2 -\-r 3 - r, when R is the radius 
of the circumcircle, r of the inscribed circle, and r v r.,, r 3 the 
radii of the escribed circles. 

(Note. — In A ABC let I be the centre of the inscribed 
circle, lj that of the escribed circle opposite A. Draw the 
diameter GDH of the circumcircle bisecting BC at D, and 
meeting the circle above BC at G below at H. Draw IM, 
IjN _L BC and produce GH to cut Mlj at K. Prove 
2 GD = r 2 -\-r 3 and 2 DH = r t - r.) 

10. If I, m, n are the _Ls from the circumcentre on the 
sides of a A, show that 

l-\-m-\-n = R + r. 
(Note. — Use the diagram and result of Ex. 9.) 



CHAPTER III 

Harmonic Ranges and Pencils 

37. A set of collinear points is called a range. 

Q8< A set of concurrent straight lines is called a pencil. 

The lines are called the rays of the pencil ; and 
their common point is called the vertex of the pencil. 

39. When three magnitudes are such that the first 
has the same ratio to the third that the difference 
between the first and second has to the difference 
between the second and third, the differences being 
taken in the same order, the magnitudes are said to 
be in harmonic proportion. (H. P.) 

Thus, if a, b and c represent three numbers such 
that a : c = b - a : c - b, a, b and c are in H. P. 

'40. If in a range of four points A, c. b, d the st. 
line AB is divided internally at C and externally at 
D in the same ratio, the distances from either end 
of the range to the other three points are in H. P. 





/ 






A 


c/ 


B 


D 


J 


/ 




jt-" 




' 




^~" 


















/ 


,---'( 














*'" 









Draw is AE, FBG to AB, making FB = BG. Draw 
EF, EG cutting AB at C, D. 



Then AC 
CB 



AE 
BF 



AE 
BG 



AD 

DB 



44 



SYNTHETIC GEOMETRY 



v 



.D 



(a) Then since AC : CB = AD : DB. 
by alternation, AC : AD = CB : DB. 

/. AC : AD = AB - AC : AD - AB. 

.*. by the definition of § 39, AC, AB, AD are in H. P. 

(b) By inversion, DB : AD = BC : AC. 
;. DB : DA = DC - DB : DA - DC. 

•'• , by § 39, DB, DC, DA are in H. P. 

State and prove a converse to this theorem. 
Gfcli When a range of four points A, C, B, D is such 
that AC : CB = AD : DB, it is called a harmonic range. 

If any point P be joined to the four points of a har- 
monic range, the joining lines form a harmonic pencil. 

Tl If, in the harmonic pencil P (A, c, B, D), a 
straight line through B parallel to PA cut PC, pd 
at E, F respectively, BE = BF. 

^P 




Fig. 44. 

V AACP||]ABCE, 
.'. AP : EB = AC : CB. 

V AADPIHABDF, 
.'. AP : BF = AD : DB. 

But, by hypothesis, AC :CB = AD : DB. 
/. AP : EB = AP : BF. 
EB = BF. 



HARMONIC RANGES AND PENCILS 45 

(43) By a proof similar to that of § 42, the following 
converse to the theorem of that article may be shown 
to be true. 

If, in the pencil p(a, c, b, D), a straight line 
through B parallel to PA cut PC, PD at E, F re- 
spectively such that be = bf, then P(a, c, b, d) is a 
harmonic pencil. 

44. Any transversal is cut harmonically by the 
rays of a harmonic pencil. 

A transversal cuts the rays PA, PC, PB, PD of the 
harmonic pencil P(A, C, B, D) at K, M, N, L respectively. 




Fia. 45. 

It is required to show that K, M, N, L is a harmonic 
range. 

Through B, N respectively draw EF, GH || PA. 

V P(A, C, B, D) is a harmonic pencil, and EBF || AP 

.-. , by § 42, EB = BF. 

™ • '1 a GN EB 1 NP BP 

From similar As, -^ = — and — = —, 

.-. , by multiplication, GN : NH = EB : BF, 

.-. GN = NH; and /., by § 43, K, M, N, L is a 

harmonic range. 



46 SYNTHETIC GEOMETRY 



middle 


[f A, 
poii 

or, 


C, B ; 

it of 


, D is a harmonic range and O is the 
AB, 

OB 2 = OC . OD. 


?c-c3 

and 


A 


O C B 

Fio. 46. 

AC AD 

CB~DB' 

AC + CB AD+DB 

AC - CB ~ AD - DB' 
20B 20 D 
20c ^OB' 
OB 2 = OC . OD. 


D 

£<S2> ^ £ 



/State and prove a converse to this theorem. 

46. If A, C, B, D is a harmonic range, A and B are 
said to be harmonic conjugates with respect to C 
and D ; and C and D are said to be harmonic con- 
jugates with respect to A and B. 

47.— Exercises 
(a) 

1. Show how to find the fourth ray of a harmonic pencil 
when three rays are given. 

2. Prove the theorem of § 44 when the transversal cuts 
the rays produced through the vertex. 

3. In the A ABC the bisectors of the interior and 
exterior ^s at A cut BC and BC produced at D, E 
respectively. Show that B, D, C, E is a harmonic range. 

s£. A, C, B, D is a harmonic range and P is any point 
on the circle described on AB as diameter. Show that- 
PA, PB respectively bisect the exterior and interior vertical 
Zs of A CPD. 

(Note. — Draw EBF II AP cutting PC, PD at E, F re- 
spectively.) 



EXERCISES 47 

5. Using Ex. 4, give geometrical proofs of the converse 
theorems of § 45. 

/6T Two circles cut orthogonally. A st. line through the 
centre of either cuts the circles at A, C, B, D. Show that 
A, C, B, D is a harmonic range. 

/K. A, C, B, D is a harmonic range. Show that the 
circles described on AB, CD as diameters cut each other 
orthogonally. 

(b) 

8. A, C, B, D is a harmonic range ; O is the middle 
point of AB and R is the middle point of CD. Show that: — 

(a) AB 3 4- CD 2 = 4 OR'; 

(b) CA. CB = CD . CO; 

(c) If P is any point on the range and lengths in 

opposite directions from P are different in sign, 
PA . PB + PC . PD = 2 PO . PR. 

Jf. The inscribed circle of A ABC touches BC, CA, AB 
at D, E, F respectively, and DF meets CA produced at P. 
Show that C, E, A, P is a harmonic range. 

(Note. — Use Menelaus' Theorem.) 

10. The diameter AB of a circle is ± to a chord CD. 
P is any point on the circumference. PC, PD cut AB, or 
AB produced, at E, F. Show that A, E, B, F is a harmonic 
range. 

11. Through E, the middle point of the side AC of the 
A ABC, a transversal is drawn to cut AB at F, BC 
produced at D, and a line through B || CA at G. Show 
that g, F, E, D is a harmonic range. 

(Note.— Use § 43.) 



48 SYNTHETIC GEOMETRY 

12. A common tangent of two given circles is divided 
harmonically by any circle which is coaxial with the given 
circles. 

(Note. — Use the converse of § 45.) 

13. In a circle AC, BD are two diameters at rt. Z.s to 
each other, and P is any point on the quadrant AD. Show 
that PA, PB, PC, PD constitute a harmonic pencil. 

14. In the harmonic pencil O (A, C, B, D), Z AOC = 
Z COB = Z BOD. Show that each of these Zs = 45°. 

15. A square is inscribed in a circle. Show that the 
pencil formed by joining any point on the circumference to 
the four vertices of the square is harmonic. 

16. P is a fixed point and OX, OY two fixed st. lines. 
Show that the locus of the harmonic conjugate of P with 
respect to the points where any st. line drawn from P cuts 
OX, OY is a st. line passing through O. 



THE COMPLETE QUADRILATERAL 



49 



The Complete Quadrilateral 

@. The figure formed by 
four straight lines which meet 
in pairs in six points is called 
a complete quadrilateral. 

The figure ABCDEF is a 
complete quadrilateral, of 
which AC, BD and EF are 
the three diagonals. 

(49, In a complete quadrilateral each diagonal is 
divided harmonically by the two other diagonals, 
and the angular points through which it passes. 

ABCDEF is a complete quadrilateral having the 
diagonal AC cut by DB at P and by EF at Q. 





It is required to show that A, P, C, Q is a harmonic 

range. 

In A ACF, AB, CD, FQ drawn from the vertices and 
meeting the opposite sides at B, D, Q are concurrent at 
E and .*. , by Ceva's Theorem, 

FD. AQ. CB = DA . QC. BF. 



*■?*.%' 



50 SYNTHETIC GEOMETRY 




The transversal DPB cuts the sides of the 


A ACF at 


D, P, B and /. , by Menelaus' Theorem, 




FD . AP . CB DA . PC . BF. 


V- 


Hence, by division, 




AQ _ Q^, 


AP PC f 


\ 


AP AQ 





or ' PC ~ QC 

and .-. A, P, C, Q, is a harmonic range. 

From the above result it is seen that F (A, P, C, Q) 
is a harmonic pencil, and consequently, by § 44, 
D, P, B, R is a harmonic range. 

Show, in the same manner that F, Q, E, R is a 
harmonic range. 



Poles and Polars 

50. If through a fixed point a line be drawn to cut 
a given circle and at the points of intersection tangents 
be drawn, the locus of the intersection of the tangents 
is called the polar of the fixed point; and the fixed 
point is called the pole of the locus. 



POLES AND POLARS 



51 



51. If c is the centre of a given circle and d is a 
fixed point, the polar of D with respect to the 
circle is a straight line which is perpendicular to 
CD and cuts it at a point E such that CE . CD equals 
the square on the radius. 




Fig. 49. Fig. 60. 

Through D draw any st. line cutting the circle at 
A and B. At A, B draw tangents to the circle inter- 
secting at P. 

P is a point on the polar of D. 
Join CD and from P draw PE J_ CD. 
Join CP cutting AB at F. Join CB. 
V Zs DEP, DFP are rt. Zs, 
.*. D, E, F, P are concyclic. 
.*. CE . CD = CF . CP. - P&C^'* 

But CF . CP = CB'. 7JP) 

.'. CE^.CD = CB'.= r i 
Then, since CD and CB^are constants, CE must also 
be constant. 

.-. the polar of D must be the st. line 1CD through 
the fixed point E such that CE . CD = CB 2 . 




52 SYNTHETIC GEOMETRY 

J>2. If a point P lies on the polar of a point Q 
with respect to a circle, then Q lies on the polar 

of P. 

P is any point on PM 

the polar of Q. 

To show that Q lies 
on the polar of P. 

Join CP and draw 
QN X CP. 

V Zs at M and N 

are rt. Zs. 

.*. Q, N, 
concyclic. 

.-. CP . CN = CM . CQ but, by § 51, CM 
square on the radius. 

.*. CP . CN = the square on the radius, and QN is 
the polar of P. 

53. Two points, as P and Q in Fig. 51, such that the 
polar of each passes through the other, are called 
conjugate points with respect to the circle, and their 
polars are called conjugate lines. 

A A such that each side is the polar of the oppo- 
site vertex is said to be self-conjugate. 



P, M are 



CQ = the 



EXERCISES 53 

54.— Exercises 

(«) 

Y? P is a point at a distance of 4 cm. from the centre 
of a circle of radius 6 cm. Construct the polar of P. 

X- P is a point at a distance of 7 cm. from the centre 
of a circle of radius 5 cm. Construct the polar of P. 

3. Draw a st. line at a distance of 7 cm. from the centre 

of a circle of radius 4 cm. Construct the pole of the line. 

/£. When the point P is within the given circle, the 

polar of P falls without the circle ; and when P is without 

the circle, the polar of P cuts the circle. 

5. The polar of a point on the circumference is the 
tangent at that point. 

9f. P is a point without a given circle and the polar of 
P cuts the circle at A. Show that PA is a tangent to the 
circle. 

Give a general statement of this theorem. 
^ If any number of points are collinear, their polars 
with respect to any circle are concurrent. 

^ Any number of lines pass through a given point; find 
the locus of their poles with respect to a given circle. 

9l If the pole of a st. line AB with respect to a circle 
is on a st. line CD, the pole of CD is on AB. 

"KL The st. line joining any two points is the polar with 
respect to a given circle of the intersection of the polars 
of the two points. 

J }A. The intersection of any two st. lines is the pole with 
respect to a given circle of the line joining the poles of 
the two st. lines. 

yi. Show how to draw any number of self -con jugate As 
with respect to a given circle. 



54 SYNTHETIC GEOMETRY 

<*) 

1/5. If a st. line PAB cut a circle at A, B and cut the 
polar of P at C, and if D be the middle point of AB, 
PA . PB = PC . PD. 

y(. Two circles ABC, ABD cut orthogonally. Show that 
the polar of D, any point on the circle ABD, with respect 
to the circle ABC passes through E, the point diametrically 
opposite to D. 

PA, The polar of any point A with respect to a given 
circle with centre O cuts OA at B. Show that any circle 
through A and B cuts the given circle orthogonally. 

r<i, A is a given point and B any point on the polar of 
A with respect to a given circle. Show that the circle 
described on AB as diameter cuts the given circle ortho- 
gonally. 

17. ABC is a A inscribed in a circle, and a || to AC 
through the pole of AB with respect to the circle meets 
BC at D. Show that AD = CD. 



55. Any straight line which passes through a 
fixed point is cut harmonically by the point, any 
circle, and the polar of the point with respect to 
the circle. 

P is the fixed point, O the centre of the circle, 
PACB any line through P cutting the circle at A, B 
and the polar EC of P with respect to the circle at C. 



POLES AND POLARS 



55 



It is required to show that B, C, A, P is a harmonic 
range. 




Join BO, OA, BE, EA and produce BE to F. 
By § 51, OP . OE = OB 2 , and .'. OP:OB = OB:OE; 
also the Z BOE is common to the A POB, BOE; 

.'. these A are similar, 
and consequently Z OEB - z OBP = Z OAB. 
.'. the points B, O, E, A are concyclic. 
.*. the Z AEP = Z OBA = Z OEB = Z FEP, and EP 
bisects the exterior vertical Z of A EBA. 

But Z PEC is a rt. Z and .'. Z BEC = Z CEA. 
BC BE BP 

" CA ~ EA ~ PA' 

and B, C, A, P is a harmonic range. 

X If the point P is within the circle, C is without the 

circle, and by § 52, the polar of C passes through p. 

.'. by the above proof B, P, A, C is a harmonic 
range. 

k Prove this theorem when the line PAB passes through 
the centre of the circle. 



56 SYNTHETIC GEOMETRY 

56. ABCD is a quadrilateral inscribed in a circle. 
AB, DC are produced to meet at E ; BC, AD to meet 
at F, forming the complete quadrilateral ABCDEF. 




AC cuts BD at G, FG cuts AB at L. 

From the complete quadrilateral FDGCAB, A, L, B, E 
and D, K, C, E are harmonic ranges, d (§49.) 

.*. L and K are points on the polar of E ; (§ 55.) 
that is, GF is the polar of E. 

Similarly, GE is the polar of F. 

Hence FE is the polar of G ; and the A EFG is self- 
conjugate with respect to the circle ABC. 

Cor.: — If from any point E two st. lines EBA, ECD are 
drawn to cut a circle at the points B, A, C, D, then 
the intersection of BD and CA is on the polar of E, and 
so also is the intersection of BC and AD. 



EXERCISES 57 

57.— Exercises 

(«) 

X. Using a ruler only, find the polar of a given point with 
respect to a given circle. 

/S. Using a ruler only, draw the tangents from a given 
external point to a given circle. 

3. Using a ruler only, find the pole of a given st. line with 
respect to a given circle. 

4. A and B are two points such that the polar of either 
with respect to a circle, with centre O, passes through the 
other. Prove that the pole of A B is the orthocentre of the 
A AOB. 

f. Tangents AB, AC are drawn to a circle. The tangent 
at any point P cuts BC, CA, AB at X, Y, Z respectively. 
Show that X, Z, P, Y is a harmonic range. 

6. If, in figure 53, ABC is a fixed A and D is a variable 
point on the circle, prove that each side of the A EFG 
passes through a fixed point. 

(Note.— Use Ex. 9, § 54.) 
Jt: C is the middle point of a chord AB of a circle, and 
D, E are two points on the circumference such that CA 
bisects the L DCE. Prove that the tangents at D and E 
intersect on AB. 

(Note. — At C draw the ± to AB and produce it to 
meet DE.) 

)£ D is the middle point of the hypotenuse BC of the 
rt.-Z-d A ABC. A circle is described to touch AD at A. 
Prove that the polar of either of the points B, C with 
respect to the circle passes through the other point. 

jtf. A, B, C, D are successive points on a st. line. Find 
points X, Y that are conjugate to each other both with 
respect to A, B and with respect to C, D. 



58 SYNTHETIC GEOMETRY 

(Note. — Draw a circle through A, B and another through 

C, D, intersecting each other at P, Q. Produce PQ to 
cut the given line at O and from O draw a tangent OT to 
either of the circles. With centre O and radius OT describe 
a circle cutting the given line at X, Y.) 

10. AD, BE, CF are concurrent st. lines drawn from the 
vertices of A ABC and cutting the opposite sides in D, E, F. 
EF meets BC at X. Show that X is the harmonic conjugate 
of D with respect to B and C. 

11. A transversal cuts the sides BC, CA, AB of the 
A ABC in D, E, F. The st. line joining A to the inter- 
section of BE and CF meets BC at H. Show that D and 
H are harmonic conjugates with respect to B and C. 

12. P, Q, are two conjugate points with respect to a 
circle. Show that the circle on PQ as diameter cuts the 
given circle orthogonally. 

13. The Z C in A ABC is obtuse and O is the ortho- 
centre. A circle described on OA as diameter cuts BC at 

D. Show that the A ABC is self-conjugate with respect to 
the circle with centre O and radius OD. 

(*) 

14. If a quadrilateral be circumscribed about a circle, the 
st. lines joining the points of contact of opposite sides are 
concurrent with the two diagonals of the quadrilateral. 

(Note. — Produce the opposite chords of contact to meet, 
and use § 56 — Cor.) 

15. If PM, QN be respectively drawn _L to the polars of 
Q, P with respect to a circle whose centre is O, PM : QN = 
OP : OQ. Salmon's Theorem. 

(Note. — Draw OK ± PM, OL _L QN ; and use similar As 
QPK, OQL.) 



EXERCISES 59 

16. D, E, F are points on the sides BC, CA, AB of the 
A ABC, such that AD, BE, CF are concurrent. Prove that 
the harmonic conjugates of D with respect to B and C, of 
E with respect to C and A, and of F with respect to A and 
B are collinear. 

17. In A ABC, X is the projection of A on BC, and BC 
is produced to cut the radical axis of the circumcircle and 
the N.-P. circle at P. Show that B, X, C, P is harmonic. 

18. The opposite sides of the quadrilateral A BCD are 
produced to meet at E, F. The diagonals of the complete 
quadrilateral form the A LMN. Show that the circle 
described on any one of the diagonals as diameter cuts the 
circumcircle of A LMN orthogonally. 



60 



SYNTHETIC GEOMETRY 



Maxima and Minima 

(58,; If a magnitude, such as the length of a st. line, 
an angle, or an area, varies continuously, subject to 
given conditions, it is said to be a maximum when it 
has its greatest possible value; and a minimum when 
it has its least possible value. 

(59) Let the distance, PA from a 
fixed point P within a circle to the 
circumference be the magnitude in 
question. 




Join A to the centre O and produce 
PO to meet the circumference at 

B and C. Afl + o# ; ^oy-0"B 

o»- - OP •. pC. 

PA < PO + OA but > OA - OP ; 

.'. PA < PB but > PC. 

As PA rotates about P its length varies continuously. 
When it comes to the position PB it is greater than 
in the positions close to PB on either side, and has its 
maximum value. 

Again, at PC it is less than in the positions close to 
PC on either side, and has its minimum value. 

Draw the diagram and illustrate in the same manner 
the maximum and minimum distances from P to the 
circumference when P is without the circle. 

What do the maximum and minimum values become 
when P is on the circumference ? 



Other simple examples are: — 



MAXIMA AND MINIMA 61 

(pi) The _L is the minimum distance from a given 
point to a given st. line; 

vj^f The minimum distance between points on two || 
st. lines is _L to the || lines ; 

jj^f The _l_ from a point on the circumference of a 
circle to a fixed chord is a maximum when the J-, or 
the ± produced, passes through the centre. 

60. (a) A and B are two fixed points on the same 
side ofa fixed st. line CD. It is required to find the 
point P in CD such that PA + PB is a minimum. 

Draw BM 1 CD and pro- ~-*;E 

duce making ME = BM. Draw ^ Q ,--^' p/ A ' | M 
AE cutting CD at P. !*^T^ ~T^ 

Then P is the required \ / '" 

point. i / 

'/A 

Take any other point Q, in 

J v Fig. 66. 

CD. Join PB, QA, QB, QE. 

AQ + QE > AE. But QE = QB and PE = PB. 

.'• AQ + QB > AP + PB ; 

and hence AP + PB is the minimum value. 

It is easily seen that AP and PB make equal Zs 
CPA and BPD with CD, and hence: — 

The sum of the distances from A and B to the st. 
line is a minimum when the distances make equal Zs 
with the line. 

Find the point P when A and B are on opposite 
sides of AB. 



62 



SYNTHETIC GEOMETRY 



D C 

x 7x .XH Y 

A <0 ^B 






(£)■ Of all As on the same base and having the same 
area the isosceles A has the least perimeter. 

Since the area is constant 
the locus of the vertices is 
a st. line XY || AB. 

If A ACB is isosceles, and 
DAB is any other A on AB 
and having its vertex in 
XY, 

Z XCA = Z CAB = Z CBA = Z YCB ; 
and .*. , by (a), AC + CB<AD+DB; 

that is, the perimeter of A ACB is less than that of 
any other A on the same base AB and having the 
same area. 

Of all As inscribed in a given acute- Zd A the 
al A has the least perimeter. 

If DEF be any A inscribed 
in ABC and FE be considered 
fixed, by (a), the sum of FD 
and DE will be least when 
Z FDB = Z EDC; thus for the 
minimum perimeter the sides 
FD, DE must make equal Zs 
with BC. Similarly DF, EF 
must make equal Zs with AB 
and DE, EF must make equal Zs with CA. 

The sides of the pedal A XYZ make equal Zs with 
the corresponding sides of A ABC. 

.*. the perimeter of XYZ is less than that of any 
other A inscribed in ABC. 





MAXIMA AND MINIMA 63 

(c^The rectangle contained by the two segments of 
a^st. line is a maximum when the st. line is bisected. 

Let P be any point in AB, and 
O the middle point. 

Describe the semicircle ACB 
and draw OC and PD _i_ AB. 

Join O, D. Fig. 58. 

AP . PB = PD 2 and AO . OB = OC 2 ; 

but V OC = OD and OD > PD .*. OC>PD; 

and /. AO . OB > AP . PB. 

(e^If the area of a rectangle is constant, its perimeter 
is a minimum when the rectangle is a square. 

In Fig. 58, rect. AP . PB = PD 2 . 

The perimeter of the square = 4 PD while the 
perimeter of the rectangle = 2 AB = 4 OD, and PD < OD. 

.'. The perimeter of the square is less than that of 
the rectangle of the same area. 

61.— Exercises 

(a) 

y. Through a given point within a given circle draw the 
chord of min. length. 

•2L A and B are two fixed points, and CD is a fixed st. 
line. Find the point P in CD, such that the difference 
between PA and PB is a maximum. 

(a) When A and B are on the same side of CD ; 

(b) When A and B are on opposite sides of CD. 

v 4fc Two sides AB, AC of a A are given in length. Show 
that the area of the A is a max. when ^ A is a rt. ^. 



64 SYNTHETIC GEOMETRY 

4. A, B are two fixed points and P is any point. Show 
that PA 2 + PB 2 is a min. when P is the middle point of 
AB. 

5. A, B, C are fixed points and P is any point. Show 
that PA 2 + PB 2 + PC 2 is a min. when P is the centroid of 
the A ABC. 

(Note.— See O.H.S. Geometry, Ex. 16, Page 133.) 

6. Find the max. and min. distances between two points 
one on each of two given non-intersecting circles. 

.7. Given two adjacent sides describe the j| gm of max. area. 

8. A, B are two fixed points. Find a point P on a fixed 
circle such that PA 2 + PB 2 is a max. or min. 

^. Of all As of given base and given vertical /, the 
isosceles A has the greatest area. 

10. Prove that the greatest rectangle that can be in- 
scribed in a given circle is a square. 

11. Give examples showing that if a magnitude vary 
continuously, there must be between any two equal values 
of the magnitude at least one maximum or minimum value. 

12. Of all chords drawn through a given point within a 
circle, that which is bisected at the point cuts off the 
min. area. 

13. From a given point without a circle, of which O is 
the centre, draw a st. line to cut the circumference in L 
and M, such that the A OLM may be a max. 

(Note. — Use Ex. 3 in the analysis of this problem.) 

14. Given two intersecting st. lines and a point wifchin 
the ^ formed by them, of all st. lines drawn through the 
point and terminated in the st. lines that which is bisected 
by it cuts off the min. area. 



EXERCISES 65 

V6. Given the base and the perimeter of a A show that 
the area is a max. when the A is isosceles. 

16. Of all As having a given area, the equilateral has 
min. perimeter. 

(Note. — Let ABC be a A having the given area and let 
two of the sides AB, AC be unequal. Then, by § 60, (b), 
if an isosceles A be described on BC of the same area, it 
will have less perimeter. .*. if any two of the sides be 
unequal the perimeter is not a min., and hence the equi- 
lateral A has the min. perimeter.) 

17. Of all rt.-Zd As on the same hypotenuse the isosceles 
A has the max. perimeter. 

^S^Find a point in a given st. line such that the sum of 
the squares of its distances from two given points is a min. 

^f. A and B are two given points on the same side of 
a given st. line; find the point in the line at .which AB 
subtends the max. Z. 

(Note. — Describe a circle to pass through the two given 
points and touch the given st. line.) 

20. Two towns are on opposite sides of a canal, unequally 
distant from it, and not opposite to each other. Where 
must a bridge be built, _L to the sides of the canal, that 
the distance between the towns, by way of the bridge, 
may be a min. % 

(*) 

21. A || gm is inscribed in a given A by drawing from 
a point in the base st. lines || to the sides. Prove that 
the area of the ||gm is a max. when the lines are drawn 
from the middle point of the base. 

22. The max. rectangle inscribed in a given A equals 
half the A. 

(Note.— Use Ex. 21.) 



66 SYNTHETIC GEOMETRY 

23. One circle is wholly within another circle, and contains 
the centre of the other. Find the max. and min. chords of 
the outer circle which touch the inner. 

24. A, B are fixed points within a given circle. Find a 
point P on the circumference such that when PA, PB 
produced meet the circumference at C, D respectively, 
CD is a max. 

(Note. — Describe a circle through A and B and touching 
the given circle.) 

25. Find the point in a given st. line from which the 
tangent drawn to a given circle is a min. 

26. Through a point of intersection A of two circles 
draw the max. st. line terminated in the two circumferences. 

(Note. — Draw CAD || the line of centres and any other 
st. line EAF. Join C, D, E, F to the other point of inter- 
section and use similar /\s.) 

27. P, Q, R are points in the sides MN, NL, LM of 
A LMN and RQ || MN. Find the position of RQ for which 
the A PQR is a max. 

(Note.— Use Ex. 21.) 

28. A is a fixed point within the L XOY. In OX, OY 
find points C, D respectively, such that the perimeter of 
the A ACD is a min. 

29. A, B are points without a given circle. On the circle 
find points P and Q such that L APB is a max. and L AQB 
is a min. 

30. The L A of the A ABC is fixed and the sum of 
AB, AC is constant. Prove that BC is a min. when 
AB = AC. 

31. A is a fixed point within the L XOY. The st. line 
BAC cuts OX, OY at B, C. Prove that BA . AC is a 
min. when OB = OC. 



EXERCISES 67 

32. From any point D in the hypotenuse BC of a rt.-^-d 
A ABC J_s DE, DF are drawn to AB, AC respectively. 
Find the position of D for which EF is a min. 

33. If the sum of the squares on two lines is given, the 
sum of the lines is a max. when they are equal. 

34. CAD is any st. line through a common point A of 
circles CAB, DAB. Prove that CA . AD is a max. when 
the tangents at C, D meet on BA produced. 

(Note. — Let E, F be the centres of circles ABD, ABC 
respectively. Join EA, and draw the radius FC II EA. Join 
CA and produce to D. Then tangents at C, D will inter- 
sect on BA. Through A draw GAH terminated in the circles. 
Prove GA. AH < CA . AD.) 

35. Describe the maximum A DEF which is similar to a 
given A ABC and has its sides EF, FD, DE passing 
respectively through fixed points P, Q, R which are not 
collinear. 

(Note. — On QR, RP describe segments containing Zs = 
Z A, Z B respectively. Through R draw a st. line || to the 
line of centres of these segments and terminated in the 
arcs at D, E. DQ, EP meet at F, giving the max. A DEF. 
In the proof use the proposition that similar As are as the 
squares on homologous sides.) 

36. A, B are fixed points on the same side of a fixed 
st. line XY. Place points P, Q on XY such that the 
distance PQ equals a given st. line and AP + BQ is 
a min. 

(Note. — Through A draw AC || XY and equal to the given 
length for PQ. Draw CM 1 XY and produce, making 
MD = CM. Join BD cutting XY at Q. Draw AP II CQ, 
cutting XY at P.) 



J 

1 

■ M , 

68 SYNTHETIC GEOMETRY 

Miscellaneous Exercises 

62. — Exercises on Loci 

(a) 

1. Construct the locus of a point such that the _l_s from 

it to two intersecting st. lines are in the ratio of two 

given st. lines. 

^ A fixed point O is joined to any point A on a given 
st. line which does not pass through O. P is a point on 
OA such that the ratio of OP to OA is constant. Find 
the locus of P. 

3. A fixed point O is joined to any point A on the 
circumference of a given circle, P is a point on OA such 
that the ratio of OP to OA is constant. Prove that the 
locus of P is a circle having its centre in the st. line 
joining O to the centre of the given circle. 

Find the locus when P is on AO produced. 

4. A fixed point O is joined to any point A on a given 
st. line which does not pass through O. P is a point on 
OA such that the rect. OP . OA is constant. Show that 
the locus of P is a circle. 

Find the locus when P is on AO produced. 
,-ST Through a fixed point O within an Z YXZ draw a 
st. line MON, terminated in the arms of the Z, and such 
that the rect. OM.ON has a given area. 

Off Find the locus of a point such that the sum of the 
squares on its distances from the arms of a given rt. Z is 
equal to the square on a given st. # line. 

7. The locus of a point, such that the sum of its distances 
from two given intersecting st. lines equals a given st. 
line, consists of the sides of a rectangle ; and the locus of 
a point such that the difference of its distances from the 
intersecting st. lines equals the given st. line, consists of 
the produced parts of the sides of the rectangle. 



EXERCISES ON LOCI 69 

8. Given the base QR of a A and the ratio of the other 
two sides, show that the locus of the vertex P is a circle 
with a diameter ST in the line QR such that S, T are 
harmonic conjugates with respect to Q and R. 

(Note. — The circle of Apollonius, see O.H.S. Geometry, 
page 235.) 

(Historical Notk.— Apollonius of Perga died in Alexandria about 200 B.C.) 

,.9.'AB is a fixed chord in a circle and C is any point on 
the circumference. Show that the loci of the middle points 
of CA, CB are two equal circles. 

10. Find the locus of the points from which tangents 
drawn to two concentric circles are _L to each other. 

11. Construct the locus of the centre of the circle of 
given radius which intercepts a chord of fixed length on a 
given st. line. 

12. Show that the locus of the centre of a circle of 
radius R which cuts a given circle at an Z A consists of 
two circles concentric with the given circle. 

13. A circle rotates about a fixed point in its circum- 
ference. Show that the locus of the points of contact of 
tangents drawn || to a fixed st. line consists of the circum- 
ferences of two circles. 

14. AB, CD are two chords of a circle, AB being fixed 
in position and CD of given length. Find the loci of the 
intersections of AD, BC and of AC, BD. 

15. A and B are the centres of two circles which intersect 
at C ; through C a st. line is drawn terminated in the 
circumferences at D and E. DA, EB are produced to meet 
at P. Find the locus of P. 

16. In a quadrilateral ABCD, AB is fixed in position, 
AC, BC and AD are given in length : — 



70 SYNTHETIC GEOMETRY 

(a) Show that the locus of P, the middle point of BD 
is a circle having its centre at E, the middle point of AB, 
and its radius equal to half of AD ; 

(b) If F is the middle point of AC, show that the locus 
of the middle point of FP is a circle having its centre at 
the middle point of FE and its radius equal to one fourth 
of AD. 

<») 

17. "What is the locus of the point P when the st. line 
MN which joins the feet of the -Ls PM, PN drawn to two 
fixed lines OX, OY is of given length. 

(Note. — Find A in OY such that AR drawn _L OX = MN. 
Draw the diameter ME of the circle through O, M, N, P ; and 
join NE. A MNE = AORA; ;. ME = OA. But OP = ME, 
.*. OP = OA and .'. the locus of P is a circle with centre O 
and radius OA.) 

18. BAC is any chord passing through a fixed point A 
within a given circle with centre E. Circles described on 
BA, AC as chords touch the given circle internally at B, C 
respectively and cut each other at D. Show that the locus 
of D is a circle described on AE as diameter. 

(Note. — F, G are centres of circles BAD, CAD respectively. 
Join FG cutting EA at H. Prove that AFEG is a || gm 
and that HD = HA.) 

19. Any secant ABD is drawn from a given point A to 
cut a given circle at B and D. Through A, B and A, D 
respectively two circles are drawn to touch the given 
circle; show that the locus of their second point of inter- 
section is a circle on the line joining A to the centre of 
the given circle as diameter. 

20. In A ABC, two circles touch AB at B and AC at 
C respectively and touch each other. Find the locus of 
their point of contact. 



EXERCISES ON LOCI 71 

(Note. — Draw BD, CD J_ respectively to BA, CA. De- 
scribe any circle with centre R in BD and passing through 
B. Produce DC to E making CE = RB. In DC find a 
point S equidistant from R and E. S is the centre of the 
circle which touches AC at C and the circle with centre R 
at P. Prove Z BPC = 180° - — and that consequently the 
locus is a segment on BC.) 

21. Any transversal cuts the sides BC, CA, AB of a 
given A ABC at D, E, F respectively. The circumscribed 
circles of the As AFE, CED cut again at P. Show that 
the locus of P is the circumcircle of A ABC. 

22. From C, any point on the arc ACB, CD is drawn 
J_ ABI with centre C and radius CD a circle is described. 
Tangents from A and B to this circle are produced to meet 
at P. Find the locus of P. 

23. Two similar As ABC, AB'C have a common vertex 
A, and the A AB'C rotates in the common plane about 
the point A. Show that the locus of the point of inter- 
section of CC and BB' is the circumscribed circle of A ABC. 

24. If a A ABC remains similar to itself while it turns 
in its plane about the fixed vertex A and the vertex B 
describes the circumference of a circle, show that the locus 
of C is a circle. 

25. AB is a fixed diameter of a given circle, E the centre 
and C any point on the circumference. Produce BC to D 
making CD = BC. Show that the locus of the point of 
intersection of AC and ED is a circle on diameter AF such 
that A and F are harmonic conjugates with respect to 
E and B. 

26. A rectangle inscribed in a given A ABC has one of 
its sides on BC. Show that the locus of the point of 
intersection of its diagonals is the line joining the middle 
point of BC to the middle point of the _|_ from A to BC. 



72 SYNTHETIC GEOMETRY 

(Note. — From the point where the median from A cuts 
the upper side of one of the rectangles draw a 1 to BC) 

27. Any chord BAC is drawn through a fixed point A 
within a circle. On BC as hypotenuse a rt.-/d A BPC is 
described such that A is the projection of P on BC. Find 
the locus of P. 

28. Any circle is drawn through the vertex of a given 
Z. Show that the loci of the ends of that diameter which 

is || to the line joining the points where the circle cuts the 
arms of the / are two fixed st. lines _L to each other and 
through the vertex of the given Z. 

29. Through C, a point of intersection of two given 
circles, a st. line ACB is drawn terminated in the circum- 
ferences at A and B. Prove that the locus of the middle 
point of AB is a circle passing through the points of inter- 
section of the given circles, and having its centre at the 
middle point of the st. line joining their centres. 

30. From a fixed point P, two st. lines PA, PB, at rt. 
Zs to each other, are drawn to cut the circumference of a 
fixed circle at A and B. Show that the locus of the middle 
point of AB is a circle having its centre at the middle 
point of the st. line joining P to the centre of the given 
circle. 

31. A || gm is inscribed in a given quadrilateral ABCD 
with sides || AC and BD. The locus of the point of inter- 
section of the diagonals of the || gm is the st. line joining 
the middle points of the diagonals of the quadrilateral. 



THEOREMS 73 

63. — Theorems 

Definition.— If A, B be the centres of two circles, and 
points P, Q be found in AB and AB produced such that 
AP AQ R 
PB ~~ QB ~~ r 
similitude of the circles. 



,, the points P, Q are called the centres of 



(a) 

X. The centres of similitude of two circles are harmonic 
conjugates with respect to the centres of the circles. 

/2. Show that each of the four common tangents of two 
circles passes through one of the centres of similitude of the 
circles. 

X- If || diameters be drawn in two circles, each of the four 
st. lines joining the ends of the diameters will pass through 
a centre of similitude of the circles. 

X. If a circle touch two fixed circles, the line joining 
the points of contact passes through a centre of similitude 
of the two circles. 

(Note. — Use Menelaus' Theorem.) 

5. The six centres of similitude of three circles lie 
three by three on four st. lines. 

y6; If two circles cut orthogonally, any diameter of one 
which cuts the other is cut harmonically by that other. 

7. In a system of coaxial circles the two limiting points 
and the points in which any one circle of the system cmts 
the line of centres form a harmonic range. 

8. Concurrent st. lines drawn from the vertices of the 
A ABC cut the opposite sides BC, CA, AB respectively at 
D, E, F. Show that 

sin ACF . sin BAD . sin CBE = sin FCB . sin DAC . sin EBA. 



74 SYNTHETIC GEOMETRY 

9. Show that the area of A ABC = j/s (s - a) (s - b) (s - c) 
where 2s = a -f b + c. 

10. Find the area of the A ABC and also the radius of 
its circuincircle, given : — 

(i) a = 65 mm., b = 70 mm., c = 75 mm.; 

(ii) a = 7 cm., 5=8 cm., c = 9 cm. 

11. If L, M, N, be the centres of the escribed circles of 
A ABC, the circumscribed circle of A ABC is the N.-P. 
circle of A LMN. 

12. In Ex. 11 if I be the centre of the inscribed circle, 
P the point where the circumscribed circle cuts IL and 
PH be _L AC, AH equals half the sum and CH half the 
difference of b and c. 

13. If O be the orthocentre of A ABC, A, B, C, O are the 

centres of the circles which touch the sides of the pedal A. 

14. CA, CB are two tangents to a circle; E is the foot 
of the J_ from B on the diameter AD; prove that CD 
bisects BE. 

(Note. — Produce DB to meet AC produced. Join AB.) 

15. The _L from the vertex of the rt. L on the hypote- 
nuse of a rt.-^-d A is a harmonic mean between the 
segments of the hypotenuse made by the point of contact 
of the inscribed circle. 

(Note. — AB is the hypotenuse of the rt.-Zd A ABC, 

p the length of the ± from C on AB. Then s - a, s-b are 

the segments of AB made by the point of contact of the 

inscribed circle. 

2 (s- a) (s-b) (b + c-a)(c + a-b) 

H.M. of segments = -^ ^ ^-= v ^ = 

° s-a + s-b 2c 

<?~a?-b* + 2ab ab 

2c =7 = ^> 



THEOREMS 75 

16. The side of a square inscribed in a A is half the 
harmonic mean between the base and the ± from the 
vertex to the base. 

17. The circumscribed centre of a "A is the orthocentre 
of the A formed by joining the middle points of its sides ; 
and the two As have a common centroid. 

18. ABC is a A. Describe a circle to touch AC at C 
and pass through B. Describe another circle to touch BC 
at B and pass through A. Let P be the second point of 
intersection of these circles. Show that Z ACP = Z CBP 
= Z BAP ; and that the circumscribed circle of A APC 
touches BA at A. Find another point Q such that Z 
QBA = Z QAC = Z QCB. 

19. O is the orthocentre of A ABC, AX, BY, CZ are the 
_Ls from A, B, C on the opposite sides, BD is a diameter 
of the circumscribed circle. Show that : — 

(a) DC = AO ; 

(6) A0 2 + BC 2 = BO 2 + CA 2 = CO 2 -f- AB 2 = the square 
on the diameter of the circumscribed circle. 

20. If a A be formed with its sides equal to AD, BE, 
CF, the medians of A ABC, the medians of the new A 
will be respectively three-fourths of the corresponding sides 
of the original A. 

(Note. — Draw FG || and = AD. Join CG. Produce FD, 
GD to cut CG, CF at K, H.) 

21. The opposite sides of a quadrilateral inscribed in a 
circle are produced to meet; show that the bisectors of the 
two Zs so formed are _L to each other. 

22. AG is a median of the A ABC. BDEF cuts AG, AC 
and the line through A || BC at D, E, F respectively. 
Show that B, D, E, F is a harmonic range. 



76 SYNTHETIC GEOMETRY 

23. If A, C, B, D be a harmonic range, show that : — 

-A--JL+-L. 

AB AC^AD 

24. Prove that the "radical axis of the inscribed circle of 
A ABC and the escribed circle which touches BC and 
AB, AC produced bisects BC. 

25. If a st. line is divided in medial section and from 
the greater segment a part is cut off equal to the less, 
show that the greater segment is divided in medial section. 

26. If a st. line is divided in medial section, the rectangle 
contained by the sum and difference of the segments is 
equal to the rectangle contained by the segments. 

27. In Figure 33, show that the centre of the circumcircle 
of A HBC lies on the circumcircle of A AHC. 

28. Show that the radius of a circle inscribed in an 
equilateral A is one-third of that of any one of the escribed 
circles. 

29. A, B, C, D and P, Q, R, S are harmonic ranges 
and AP, BQ, CR are concurrent at a point O, Prove that 
DS passes through O. 

30. A, B, C, D and A, E, F, G are harmonic ranges 
on two st. lines AD, AG. Prove that BE, CF, DG are 
concurrent. 

P) 

31. Three circles pass through two given points P, Q. 
Two st. lines drawn from P cut the circumferences again 
at R, S, T and R', S', T. Show that RS : ST = R'S' : S'T'. 

(Note. — Join the six pcints to Q.) 

32. The middle points of the diagonals of a complete 
quadrilateral are collinear. 



THEOREMS 



ABCDEF is a complete quadrilateral; L, M, N the middle 
points of its diagonals. 

Draw LHK || AE, produce KM to cut AB at G and join 
GH. 

Prove L, M, N collinear. 

k F 




33. ABCD is a quadrilateral and O is a poi 
such that A AOB + A COD = A BOC + A 
that the locus of O is the st. line joining 
points of the diagonals AC and BD. 

(Note. — Produce DA. CB 
to meet at E. Make EF = 
AD and EG = BC. Join 
FO, EO, GO, FG.) 

A OEF = A OAD, A OEG = 
A OBC. ;. OF EG - half 
ABCD, and A EFG is con- 
stant; .'. A OFG is constant 
and is on the constant base 
FG ; ,\ the locus of O is a 

Fio. 60. 

It is easily seen that the 
middle points of the diagonal are on the locus. 



nt \v 
AOD 

the 



ithin it 
; show 
middle 




78 SYNTHETIC GEOMETRY 

34. If a quadrilateral be circumscribed about a circle, 
the centre of the circle is in the st. line joining the middle 
points of the diagonals. 

(Note.— Use Ex. 33.) 

35. G is a fixed point in the base BC of the A ABC and 
O is a point within the A such that A AOB + A COG = 
A AOC + A BOG ; show that the locus of O is the st. line 
joining the middle points of BC and AG. 

(Note.— From CB cut off CD = BG. Join OD, AD.) 

36. G is the point of contact of the inscribed circle of 
A ABC with BC. It is required to show that the centre 
of the circle is in the st. line joining the middle points of 
BC and AG. 

37. A is a fixed point on a given circle and P is a 
variable point on the circle. Q is taken on AP produced 
so that AQ : AP is constant. Show that the locus of Q is 
a circle which touches the given circle at A. 

38. S is a centre of similitude of two circles PQT P'Q'T', 
and a variable line through S cuts the circles at the 
corresponding points P, P' ; Q, Q'. Prove that, if STT' is 
a common tangent 

SP . SQ' = SP' . SQ = ST . ST'. 

39. AD is a median of the A ABC. A st. line CLM 
cuts AD at L and AB at M. Prove that ML : LC = 
AM : AB. 

40. The locus of the point at which two given circles 
subtend equal Ls is the circle described on the join of 
their centres of similitude as diameter. 

41. Having the N.-P. circle and one vertex of a A 
given, prove that the locus of its orthocentre is a circle. 

(Note. — Let A be the given vertex and K the centre 
of the N.-P. circle. Join AK and produce to M making 
KM = AK. M is the centre of the locus.) 



THEOREMS 79 

42. AB is a chord of a circle and the tangents at A, B 
meet at C. From any point P on the circle _Ls PX, PY, 
PZ are drawn to BC, CA, AB respectively. Prove that 
PX . PY = PZ 2 . 

43. OX, OY are two fixed st. lines and from them equal 
successive segments are cut off; AC, CE, etc., on OX; BD 
DF, etc., on OY. Show that the middle points of AB, CD, 
EF, etc., lie on a st. line || to the bisector of the Z XOY. 

(Note. — Produce the line joining the middle points of 
AB, CD to cut OX, OY and use Menelaus' Theorem.) 

44. Show that the st. line joining the vertex of a A to 
the point of contact of the escribed circle with the base 
passes through that point of the inscribed circle which is 
farthest from the base. 

(Note. — Show that the vertex and the point where the 
bisector of the vertical Z cuts the base are harmonic 
conjugates with respect to the inscribed and escribed 
centres ; and use the resulting harmonic pencil having its 
vertex at the point of contact of the escribed circle with 
the base.) 

Prom Ex. 44 show that the _L from the vertex to the 
base of a A and the bisector of the vertical Z are harmonic 
conjugates with respect to the lines joining the vertex to 
the points of contact of the inscribed and escribed circles 
with the base. 

45. The five diagonals of a regular pentagon intersect at 

five points within it. Show that the area of the pentagon 

7 — 3 /5~ 
with these points for vertices is ~ A, where A is 

the area of the given pentagon. 

46. ABCD is a rectangle. If A, P, C, Q and B, R, D, S 
are each harmonic ranges, show that P, Q, R, S are 
concyclic. 



80 SYNTHETIC GEOMETRY 

47. If one pair of opposite sides of a cyclic quadrilateral 
when produced intersect at a fixed point, prove that the 
other pair when produced intersect on a fixed st. line. 

What is the connection between the fixed point and the 
fixed st. line 1 

48. Concurrent st. lines drawn from the vertices of the 
A ABC cut the opposite sides BC, CA, AB respectively at 
D, E, F. Prove that the st. lines drawn through the 
middle points of BC, CA, AB respectively || to AD, BE, CF 
are concurrent. 

49. The sides AB, BC, CD, DA of a quadrilateral touch 
a circle at E, F, G, H respectively. Show that the opposite 
vertices of ABCD, the intersection of the diagonals of 
EFGH, and the intersections of the opposite sides of EFGH 
form two sets of collinear points. 

50. The circle APQ touches the circle ABC internally at 
A. The chord BC of the circle ABC is tangent to the 
circle APQ at R, and the chords AB, AC intersect the 
circle APQ in the points P, Q. Prove that AP . RC = 
AQ . BR. 

51. Tangents to the circumcircle of A ABC, at the 
vertices, meet at D, E, F. AD, BE, CF are concurrent at 
O. Show that the _Ls from O on the sides of A ABC are 
proportional to the sides. 

(Note. — Draw GDH H FE and meeting AB, AC produced 
at GH. Draw DR, DS _L AG, AH. Prove DG = DH ; and, 

OM DR AH AB \ 
if OM, ON are J. AB, AC, that o~N = DS = AG = AC°J 

If AD cuts BC at K, show that BK : KC = AB 2 : AC 2 . 

52. O is the middle point of a chord AB of a circle, DE, 
FG are any chords through O; EF, GD cut AB at H, K. 
To prove OK = OH. 



THEOREMS 81 

Produce EF, GD to meet at L; EG, FD at M. Produce 
ED to meet LM at P. Join OL, OM. 




m^:'' _ i..:E.l\ L 



OLM is a self-conjugate A ; .'. CO produced cuts LM at 
rt. ^s, and .*. AB il LM. PDOE is a harmonic range, and 
.". L (P, D, O, E) is a harmonic pencil. Then V AB through 
O is || LM and cuts the other two rays at H, K; .*. OH = 
OK. 

53. Through any point P in the median AD of A ABC a 
st. line is drawn cutting AB, AC at Q, R. Prove that 
PQ : PR = AC . AQ : AB . AR. 

(Note. —Produce BP, CP to cut AC, AB at K, L. Join 
3R, CQ. BD . CL . PK = DC . LP . KB, 



LP PK 
CL KB' 



A APQ = AAPR . A APQ = A ACQ , 
•'• A ACQ AABR' ° r AAPR A ABR 
PQ : PR = AC . AQ : AB . AR.) 



82 SYNTHETIC GEOMETRY 

64. — Problems 
(a) 

1. Draw a st. line, terminated in the circumferences of 
two given circles, equal in length to a given st. line, and 
|| to a given st. line. 

2. Through a given point on the circumference of a circle 
draw a chord which shall be bisected by a given chord. 

3. In the hypotenuse of a rt.- Z d A find a point such that the 
sum of the ±s on the arms of the rt. Z equals a given st. line. 
What are the limits to the length of the given st. line? 

4. In the hypotenuse of a rt.-Zd A find a point such 
that the difference of the -Ls on the arms of the rt. Z 
equals a given st. line. 

"When will there be two, one or no solutions 1 

5. In the hypotenuse of a rt.-Zd A find a point such 
that the ±s on the arms of the rt. Z are in a given ratio. 

6. Through a given point draw a st. line terminated in 
the circumferences of two given circles and divided at the 
given point in a given ratio. 

(Analysis. — Let A be the given point and suppose PAQ 
to be the required st. line terminated at P, Q in the circles 
of which the centres are respectively C, D. Join CP, and 
draw QR || CP meeting CA in R. From similar As, 

CA CP PA 

— — = — - = the given ratio ^TTy wherein CA } CP are 

known; and .*. the position of R and the length of QR 
are known.) 

7. In a given circle inscribe a rectangle having its 
perimeter equal to a given st. line. 

8. In a given circle inscribe a rectangle having the 
difference between adjacent sides equal to a given st. line. 

9. In a given circle inscribe a rectangle having its sides 
in a given ratio. 



PROBLEMS 83 

10. In a circle of radius 5 cm. inscribe a rectangle having 
its area 22 sq. cm. 

(Analysis. — Let x, y be the sides of the rectangle, then 
xy = 22 and x 2 + y 2 = 100. Show that x + y = 12.) 

11. A and B are fixed points on the circumference of a 
given circle. Find a point C on the circumference such 
that CA, CB intercept a given length on a fixed chord. 

(Analysis. — Draw BL_ || to the given chord and equal to 
the given length. Join L to the point where AC is supposed 
to cut the given chord. Join AL, etc.) 

12. A and B are fixed points on a circumference. Find 
a point C on the circumference such that CA, CB cut a 
fixed diameter at points equally distant from the centre. 

(Analysis. — Draw the diameter AOD. Join D to the 
point F where BC is supposed to cut the given diameter. 
Prove that the circumcircle of A DFB touches the given st. 
line AD at D.) 

13. In a given circle inscribe a A, such that two of its sides 
pass through given points, and the third side is a maximum. 

14. Two towns are on different sides of a straight canal, 
at unequal distances from it, and not opposite to each 
other. Where must a bridge be built _L to the direction 
of the canal so that the towns may be equally distant from 
the bridge 1 

15. Divide a given st. line into two parts so that the 
squares on the two parts are in the ratio of two given st. 
lines. 

16. Construct the locus of a point the difference of the 
squares of whose distances from two points 3 inches apart 
is 5 sq. inches. 

17. Two points A and B are four inches apart. Con- 
struct the locus of the point the sum of the squares of 
whose distances from A and B is 20'5 square inches. 



84 SYNTHETIC GEOMETRY 

18. Divide a given st. line into two parts such that the 
sum of the squares on the whole st. line and on one part 
is twice the square on the other part. 

(Analysis.— Let AB (= a) 
Ct^-.^^ be the required line and 

N v x "*-«. EB (= cc) a segment such that 

iA %N y/ ""*V^. B " 2 + & = 2(a - x)\ Then 



D E cc = 2a - aj/3, and a - x — 

Fig. 62. ,5- . AE \/3 - 1 _ 

ay8 -a; .. ^ = ^VS - 

1/3 + 1. .'. AE = EB |/3+ EB. Cut off ED = EB and draw 
AC J_ AB and = EB. Then since AD = EB ]/3 = AC v/3, 
Z ADC = 30° and CD = 2AC = DB. .'. Z B = 15°. 

The following construction is then evident : — At B make 
Z ABC = 15°, and draw AC _|_ AB. Draw the rt. -bisector 
of BC cutting AB at D. Bisect DB at E.) 

19. Two non-intersecting circles have their centres at A 
and B, and C is a point in AB. Draw a circle through 
the point C and coaxial with the two given circles. 

(Note. — From O, the point where the radical axis cuts 
AB draw a tangent, OT, to one of the given circles. Then 
if CC is the diameter of the required circle, OC . OC = 
OT 2 and .*. OC is the third proportional to OC and OT.) 

20. Construct a A having one side and two medians 
equal to three given st. lines. (Two cases.) 

21. Construct a A having the three medians equal to 
three given st. lines. 

22. Given the vertical Z, the ratio of the sides containing 
it, and the diameter of the circumscribing circle; construct 
the A. 

23. Given the feet of the -Ls drawn from the vertices of 
a A to the opposite sides; construct the A. 



PROBLEMS 



85 



24. Draw a circle to touch a given circle, and also to 
touch a given st. line at a given point. 

25. Draw a circle to pass through two given points and 
touch a given circle. 

26. Draw a circle to pass through a given point and 
touch two given intersecting st. lines. 

27. AB is the chord of a given segment of a circle. 
Find a point P on the arc such that AP + BP is a 
maximum. 

28. Find a point O, within a A ABC such that: — 

(1) A AOB : A BOC : A COA -1:2:3; 

(2) A AOB : A BOC : A COA = I : m : n. 



(b) 

29. Find a point such that its distances from the three 
sides of a A may be proportional to three given st. lines. 

(Note. — Draw BD, CE J_ 
BC, and each = I. Join DE. 
Draw AF ± AC, and = m. 
Draw FG II AC, meeting DE 
at G. Draw AH _L AB, and 
= n. Draw HK || AB, meet- 
ing DE at K. BK, CG meet 
at P the required point. 
Show that, if PL, PM, PN 
be ± to the sides, PL : PM 

30. Through a given point within a circle draw a chord 
which shall be divided in a given ratio at the given point. 

31. A, B, C, D are points in a st. line. Find a point 
at which AB, BC, CD subtend equal ^s. 

(Note. — The required point is the intersection ot two 
circles of Apollonius. See O.H.S. Geometry, page 235.) 




86 



SYNTHETIC GEOMETRY 



32. Given a vertex, the orthocentre and the centre of 
the N.-P. circle of a A, construct the A. 

33. Having divided a st. line internally in medial section, 
find the point of external division in medial section hy ratio 
and proportion. 

3i. Describe an equilateral A with one vertex at a given 
point, and the other two vertices on two given || st. lines. 

(Analysis. — Describe a 
circle about the equilateral 
A ABC cutting the || lines 
again at L, M. Then 
Z ALC = ABC = G0°, and 
Z AMB = Z ACB = 60°.) 
35. Find the locus of 
the middle point of the 
chord of contact of tan- 
gents drawn from a point 
on a given st. line to a 
Fia 64 - given circle. 

36. The locus of the centre of a circle which bisects the 
circumferences of two given circles is a st. line _l_ to the 
line of centres and at the same distance from the centre of 
one circle that the radical axis is from the centre of the other. 

37. Describe a circle to bisect the circumferences of three 
given circles. 

38. Find the locus of the centre of a circle that passes 
through a given point and also bisects the circumference of 
a given circle. 

39. Describe a circle to pass through two given points 
and bisect the circumference of a given circle. 

40. Given a point and a st. line, construct the circle to 
which they are pole and polar, and which passes through a 
given point. 




Part II.— ANALYTICAL GEOMETRY 



FORMULA 



The following important results from algebra and trigonometry 
are frequently used in analytical geometry : — 

1. The roots of the quadratic equations ax 2 + 2bx + c = ara 

- 6 + jW~^ and - b - ,jW^ t 

a a 

These roots are 

real, if b 2 >, or =, aa ; 

imaginary, if b 2 < ac ; 
equal to each other, if 6 2 = ac ; 
equal in magnitude but 

ODDosite in sign, if b = ; 

xationai, if b- - ac is a perfect square. 

One root = 0, if c = ; 

ooth roots = 0, if b = c = 0. 

The sum of the roots = — 

a 

The product of the roots = — . 
o 

2. The fraction ^ = oo, if 6 = and a is not = 0. 

o 

3. The equation ax + by + c = is the same as px + qy + r = 0, 

., a b c 
if — = _ = _. 

p q r 

4. ax 2 + 2bx + c is a perfect square, if b 2 = ac. 

5. If a x x + b x y + c,_z = 
and a 2 x + b 2 y + c 2 z = 0, 

then * = t. 



b l c 2 - b 2 c 1 c 1 a 2 - c 2 a 1 a 1 6 2 — a 2 b r 



V F0RMULJ5 

6. For all values of «, 

sin 2 a + cos 2 a = 1. 

7. If tan a = k, a = tan ~ l k. 

8. sin (A ± B) = sin A cos B ± cos A sin B. 

cos (A ± B) = cos A cos 3 + sin A sin B. 

9. sin A + smB = 2 sin ~*~ , cos ^-1 — , etc 

in i tK j. q\ ton A ± tan B 

10. ton (A ± B) = _— — -. 

1 + tan A ton B 



CONTENTS 



Chapter I ?Aea 

Cartesian Coordinates , 1 

Rectangular Coordinates „ . 2 

The Distance Between Two Points 6 

Area of a Triangle 14 

Loci 18 

Chapter II 

The Straight Line 25 

The Angle Between Two Straight Lines ... 40 

Perpendiculars ... 45 

Chapter III 

The Straight Line Continued 57 

Transformation of Coordinates 62 

Review Exercises . . .' 68 

Chapter IV 

The Circle . . . 72 

Tangents 79 

Poles and Polars 88 

Tangents from an Outside Point ..... 95 

Radical Axis , . . 97 

Miscellaneous Exercises 99 

Answers . . ......... 113 



ELEMENTARY ANALYTICAL GEOMETRY 

CHAPTER I 

Cartesian Coordinates 

1. Analytical, or algebraic, geometry was invented 
by Descartes in 1637, and this invention marks the 
beginning of the history of the modern period of 
mathematics. It differs from pure geometry in that 
it lays down a general method, in which, by a few 
simple rules, any property can be at once proved or 
disproved, while in the latter each problem requires a 
special method of its own. 

2. The Origin. In plane analytical geometry the 
positions of all points in the plane are determined by 
their distances and directions as measured from a fixed 
point. 

If the points are all in a st. line, the fixed point is 
most conveniently taken in that line. 



if— 3 c 1 

Fig. 1. 

Thus, if the distance and direction of each of the 
points A, B, C, D, E from the point O are given, the 
positions of these points are known. 

The point O is called the origin, or pole. 
1 



2 ELEMENTARY ANALYTICAL GEOMETRY 

3. Use of plus and minus. In algebra the signs 
plus and minus are used to indicate opposite qualities 
of the numbers to which they are prefixed ; and in 
analytical geometry, as in trigonometry, these signs are 
used to show difference of direction. In a horizontal 
st. line distances measured from the origin to the 
right are taken to be positive, while those to the left 
are negative ; and in a vertical st. line distances 
measured upward are positive, while those measured 
downward are negative. Thus, in Fig. 1, if OA = 2 cm., 
OB = 3 cm., OC = 5 cm., OD = 1 cm., and OE = 3 cm., 
the positions of these points are respectively repre- 
sented by 2, 3, 5, — 1 and — 3, the understood unit 
being one centimetre. 



Rectangular Coordinates 

4. Coordinates. When points are not in the same 
st. line, their positions are determined by their distances 



from two st. lines xOx and y'Oy drawn through the 
origin, the distances being measured in directions || to 
the given st. lines. 



RECTANGULAR COORDINATES 6 

These lines are called the axes of coordinates, or 
shortly, the axes. 

x'Ox is called the axis of x, and y'Oy is called the 
axis of y. 

From a point P draw PM i| Oy and PN || Ox, terminated 
in the axes. 

PM is called the ordinate of P, and PN ( = OM), is 
called the abscissa of P. These two distances, the 
abscissa and ordinate, are called the coordinates of 
the point. 

Sometimes, from the name of the inventor, they are 
spoken of as cartesian coordinates. 

5. Rectangular coordinates. When the axes are at 
rt. Zs to each other, the distances of a point from 
the axes are called its rectangular coordinates. 



T P 



I I I 



Fia. 3. 



To locate the point of which the abscissa is 4 and 
the ordinate 3 when the coordinates are rectangular, 



4 ELEMENTARY ANALYTICAL GEOMETRY 

measure the distance OM = 4 units along Ox and at 
M erect the J_ PM = 3 units. P is the required point. 



















































































































f! 




















































































































































































































































































































































































1 














































































































































































































































































































































/ 












































t\ 


/\ 
































































































































































































































































































































I 
































1 
































































'' 


































1 























































Fig. 4. (Unit = J inch. ) 

In Fig. 4, the abscissa of P = OM = 2 8, the ordinate 
of P = PM = 2. The position of this point is then 
indicated by the notation (2*8, 2). For Q, the abscissa = 
ON = - 1'6, the ordinate = QN = 26 and the position 
of the point is indicated by ( — 1*6, 2 - 6). 

Similarly the position of R is ( — 1, — 1'6), and 
that of S is (1-4, -12). 

xOy, yOx', x'Oy' and y'Ox are respectively called 
the first, second, third and fourth quadrants; and we 
see from the diagram, that: — 

for a point in the first quadrant both coordinates 
are positive; 



EXERCISES 5 

for a point in the second quadrant the abscissa is 
negative and the ordinate is positive ; 

for a point in the third quadrant both are negative; 
and 

for a point in the fourth the abscissa is positive and 
the ordinate is negative. 

Thus the signs of the coordinates show at once in 
which quadrant the point is located. 

6.— Exercises 

A^ Write down the coordinates of the points A, B, C, D, 
E, F, G, H and O in Fig. 5. 

















































































f t\ 


~? 


^i\r- 


























































































































\l) 




1 


. 






























*> 


















* 








































.. 














































































■s 


/ \ i 
















































\> 




























■F 




































































(Z 













































Fig. 5. (Unit = ^ inch.) 

' J£. Draw a diagram on squared paper, and mark on it 
the following points :— A (4, 3), B (4-6, 0), C ( - 2, - 3), 
D ( - 4, 2), E (0, 2-8). Indicate the unit of measurement 
on the diagram. 

&? Draw a diagram on squared paper and mark the 
following points :— (4, 3), (3, 4), ( - 3, - 4), ( - 4, 3), (0, 5), 
(0, - 5), ( - 5, 0). Describe a circle with centre O and 



6 



ELEMENTARY ANALYTICAL GEOMETRY 



radius 5. Should the circle pass through the seven points'! 
Why? 

4. The side of an equilateral A = 2a. One vertex is at 
the origin, one side is on the axis of x and the A is in the 
first quadrant. What are the coordinates of the three 

vertices 1 

^ 5. One corner of a square is taken as origin and the axes 
coincide with two sides. The length of a side is b. What 
are the coordinates of the corners, the square being in the 
first quadrant 1 ? 



The Distance Between Two Points 
7. In general, the abscissa of a point is represented 
by x, the ordinate by y. 

(C)To find the distance between a point P (x v yj 
and the origin. 



y' 

Fiq. 6. 



From P draw PM j_ Ox. 

V PMO is a rt.-Zd A, 
.-. PO 2 = OM ! -f PM 2 

/. po = ^ + y x 2 . 



THE DISTANCE BETWEEN TWO POINTS 



(9JT0 find the distance between P (x v y x ) and 




Draw PM and QN j_ Ox ; QL j_ PM. 

QL = NM = OM - ON = x l - X.,. 

PL = PM - LM = PM - QN = y l - y 2 . 

V PLQ is a rt.-Zd A, 

.*. PQ 2 = QL 2 + PL 2 . 

= (x l - x 2 y + ( Vl - yj*. 



.-. PQ = v /( Xl - x. 2 y + ( 7l - y 2 )l 

10. If the point Q in § 9 coincides with the origin 
O, x., = and y. 2 = 0. Substituting these values, in the 
expression for PQ in that article we obtain 



PO = yx* + y x \ 

This shows that the result in §8 is a particular case 
of that in § 9. 

11. The result in § 9 holds good, in the same form, 
for any two points whether the coordinates are positive 
or negative. 



8 



ELEMENTARY ANALYTICAL GEOMETRY 



For example — it is required to find the distance 
between P (- 3, 2) and Q (5, - 2). 







































































































































































































































*N 


























































































































































































N s 














































k 
































































■ 








































































































































































^s 




























































































s s 


<> 



















































Fia. 8. (Unit = ^ inch.) 

Draw PM, QN j. Ox ; QL 1 PM. 

The length of ML = length of NQ = 2. 

.-. PL = PM + ML = 2 + 2 = 4. 
The length of QL = NM = 5 + 3 = 8. 
PQ^ = ql 2 + PL 2 
= 64+ 16 = 80. 
.'. PQ = 4 y'W. 

If in the expression for PQ found in § 9, we sub- 
stitute - 3 for x v 2 for y v 5 for x 2 and - 2 for y 2 , 
we obtain 



PQ. i/(-3-5)2 + (2 + 2)s 

= 4j/5, 
the same result. 



THE DISTANCE BETWEEN TWO POINTS 9 

ff2j)The particular cases in § § 10 and 11 illustrate 
what is known as the continuity of the formulae 
in analytical geometry. Here continuity means, that 
general results which are obtained when the coordinates 
in the diagram used are all positive hold true in the 
same form for all points. 

13. To find the coordinates of the middle point 
of the distance between two given points P (x v y x ) 
and Q (x 2 , y 2 ). 



y 


Qf*-- 


JpC-_JT 

1 i 

— - js 1 

1 1 







N 


L M 


X 



Let R (x, y) be the middle point of PQ. 
Draw PM, QN, RL ± Ox ; QS j_ RL; RT 1 PM. 
From the equality of As PRT, RQS, 
QS = RT and RS = PT. 
.*. NL = LM, 

• • \JU ~~ \fj(y —— JO-t JOm 

x x 4- x. 2 



x = 
V RS = PT, 

.*. y-v*-- 



Vx - y- 



y = 



2/i + y 2 



10 ELEMENTARY ANALYTICAL GEOMETRY 

Thus the coordinates of R are 

*i + x 2 7i + y 2 

2 ' 2 ' 

Qj) To find the coordinates of the point dividing 
the distance between P (x 1 y x ) and Q (.r 2 y 2 ) in the 
ratio of m to n. 




Fia. 10. 

Let R (x, y) be the point dividing PQ such that 
PR m 
RQ — n 

Draw PM, QN, RL J. Ox ; QS J_ RL; RT ± PM. 

From the similar As PRT, RQS 

RT _ PT P R m 
QS - RS RQ -■ 



RT 
QS 



a? — cc„ 



m 
-n. ' 
?n 

71 

mx. 





EXERCISES 


li 


.". 


nx, + mx 9 
x — 

m + n 




V 


PT_ m 

RS ,7' 




••• 


2/1-2/ m 

2/ - 2/2 » 






my - my. 2 = ny 1 

y = ^2/i + my 2 
m + 7i 


- ^y. 


Thus the coord 


mates of R are 

tj + mx. 2 ny! + my 2 


-f Tx*» frTv^-flVu 


113 


>v /U." 



m + n m + n 

15. If the point R be taken in PQ produced such that 
PR : RQ = m : n, and the coordinates of P, Q be 
( x v 2/i)' ( x 2> 2/2) ik may be shown by a proof similar to 
that in the previous article that the coordinates of R are 

mx 2 - nXj my 2 - ny l 
m - n ' m - n 

These results and also those of § § 13 and 14- are 
the same for oblique and rectangular axes. 

16.— Exercises 

1. Find the distance between the points (G, 5) and (1, - 7) 
and test your result by measurement on squared paper. 

' 2. Find the distance between the points (2, - 3) and 
(-1, 1) and test your result by measurement on squared 
paper. 

3. Find the coordinates of the middle points of the st. lines 
joining the pairs of points in exercises 1 and 2 respectively 
and test the results by measurements on the diagrams. 



12 ELEMENTARY ANALYTICAL GEOMETRY 

4. Find, to two decimal places, the distance between 
( - 3, 7) and (4, - 4). 

5. The vertices of a A are ( - 2, 4), ( - 8, - 4) and (7, 4). 
Find the lengths of its sides. 

6. The vertices of a A are ( - 1, 5), ( - 4, - 2), (5, - 3). 
Find (a) the lengths of the sides ; (b) the lengths of the 
medians. 

7. The vertices of a quadrilateral are (4, 3), ( - 5, 2), 
( - 3, - 4), (6, - 2). Find the lengths of its sides, and also 
of its diagonals. 

v 8. Find the coordinates of the middle point of the st. 
line joining (3, - 2) and ( - 3, 2). 

9. Find the points of trisection of the st. line joining 
(1, 3) and (6, 1). 

10. The st. line joining P ( - 4, - 3) and Q (6, - 1) 
is divided at R (x, y) so that PR : RQ =5:2. Show that 
x = -2y. 

11. Find the length of the st. line joining the origin to 
(a, - b). 

12. The st. line joining the origin to P (-4, 7) is 
divided at R, Q so that OR : RQ : QP = 3 : 4 : 2. Find 
the distance RQ. 

13. The length of a st. line is 17 and the coordinates of 
one end are ( - 5, - 8). If the ordinate of the other end 
is 7, find its abscissa. 

J 14. Find in its simplest form the equation which expresses 
the fact that (x, y) is equidistant from (5, 2) and (3, 7). 

15. Find the centre and radius of the circle which passes 
through (5, 2), (3, 7) and ( - 2, 4). 



EXERCISES 13 

]%>. Find the points which are distant 15 from (-2, - 10) 
and 13 from (2, 14). 

v ^7. Prove that the vertices of a rt.-Zd A are equidistant 
from the middle point of the hypotenuse. 

Suggestion : — Take the vertex of the rt. Z for origin and 
the sides which contain the rt. Z for axes. 

- ISC' In any A ABC prove that 

AB 2 + AC 2 = 2 (AD 2 -f DC 2 ), 

where D is the middle point of BC. 

Suggestion : — Take D as origin, DC as axis of x and the 
J_ to BC at D as axis of y. Let DC = a, and the coordi- 
nates of A be (x v y x ). 

19. If D is a point in the base BC of a A ABC such 
that BD : DC = m : n, show that 

n AB J + m AC 2 = (m + n) AD 2 + n BD 2 + m DC 2 . 

Suggestion : — Take D as origin, DC as axis of x and the 
J_ to BC at D as axis of y. Let BD = - ma, DC = na, and 
the coordinates of A. be (x v y^. 

20. The vertices of a A are the points (x v y^), (x 2 , y 2 ), 
(tc 3 , y 3 ). Find the coordinates of its centroid. 

v 21. The st. line joining A (2, 1) to B (5, 9) is produced 
to C so that AC : BC = 7:2. Find the coordinates of C. 

22. The st. line joining A (3, - 2) to B ( - 4, - 6) is 
produced to C so that AC : BC = 3 : 2. Find the coordi- 
natas of C. 



14 



ELEMENTARY ANALYTICAL GEOMETRY 



The Area of a Triangle 

17. To find the area of the A of which the 
vertices are A (x v y^ B (.»■.,, y 2 ) and c (.r 3 , y. 6 ). 




Fio. 11. 

Draw the ordinates AL, BM, CN. 

From the diagram, 

A ABC = ALNC + CNMB - ALMB. 

The area of a quadrilateral of which two sides 
are || = half the sum of the || sides X the distance 
between the j| sides. 

.'. ALNC = £ (AL + CN) X LN = \ (y, + y s ) (x s - x x ), 
CNMB = h (CN + BM) XNM = |(i/ 3 4 2/ 2 ) ( x 2 ~ x z)> 
ALMB = h (AL 4- BM) X LM = h (y 1 + 2/o) 0' 2 - X x ). 
;. A ABC = i { (y 1 4- y z ) (x 3 - x,) + (2/3 + y 2 ) (x. 2 - x z ) - 

(3/1 + 2/2) (®2 -«l)}- 

Simplifying, 

A ABC = £ { Xl (y 2 - y s ) 4- x, (y s - y x ) 4- x 3 (y x - y 2 )} • 



THE AREA OF A TRIANGLE 



15 



Note. — The points have been taken in circular order about 
the A in the opposite direction to that in which the hands of 
a clock rotate; if they are taken in the same direction as the 
hands rotate, the formula will give the same result only it 
will appear to be negative ; but, of course, the area of a A 
must be positive. 

18. To find the area of the A of which the vertices are 
(3, 2), (-4,3), (-2, -4). 




Fig. 12. (Unit = , 3 „ inch.) 

Draw the diagram on squared paper. Draw the 
ordinates AL, BM, CN. Through C draw RCS || Ox to 
meet AL, BM produced at S, R. 

= BRSA 

91 

2 



A ABC 
BRSA 



A BRC - A ACS. 
l(BR + AS)RS=i(7 + 6)X7 



16 ELEMENTARY ANALYTICAL GEOMETRY 

14 
ABRC = |BRXRC = |X7X2 =— • 

u 

30 
AASC =|ASXSC = £X6X5 = — 

, A ABC = 91-14-30 = 47 
^ 2 2 

If we substitute the coordinates of A, B and C in the 
formula of § 17, we obtain 

AABC= i {3(3 + 4) + (-4)(-4- 2) + (- 2)(2-3)} 

47 
= 1(21 + 24 + 2)=-^; 

the same result as before. 

This illustrates the continuity of the symmetrical 
result found in § 17 for the area of a A. 

19. — Exercises 

v Q) Find, from a diagram, the area of the A of which 
the vertices are (o, o), (a, b), (c, d). Check your result by 
using the formula of § 17. 

2. Draw the following As on squared paper and find 
their areas; checking your results by using the formula of 
§ 17:— 

J& (1,4), (-2,2), (5, -1); 

(b) (4, -2), (-5,-1), (-2, -6); 

(c) (0,0), (3,4-5), (-2-5,4). 

3. Find the area of the quadrilateral of which thtf 
vertices are (3, 6), ( - 2, 4), (2, - 2) and (7, 3). 

4. Find the area of the quadrilateral of which the 
vertices are (0, 0\, (4, 0), (3, 6) and ( - 3, 3). 



. THE AREA OF A TRIANGLE 17 

^ 5. D, E, F are respectively the middle points of the sides 
BC, CA, AB of a A. Prove by the formula of § 17, 
taking B as origin and BC as axis of x, that A ABC - 4 A 
DEF. 

^NL Find the area of the A of which the vertices are 
(x, y), (3, 5), (-2, 4); and thence show that if these 
points are in a st. line 5y - x = 22. 

&L Find the area of the A A ( - 3, 2), B (7, 2), C (3, 10); 
and show that the J_ from A to BC = BC. 

V 'S^ A man starts from O and goes to A, from A to B, 
B to C, C to D, D to O. If O be taken as the origin 
and the coordinates of A, B, C, D are (0, - 3), (8, 3), 
(-4, 8), (-4, 3), find the distance he has travelled, the 
unit being one mile. 

'^^ Show from the formula for the area of a A that 
A (3, -2), B (19, 10) and C (7, 1) are in the same st. 
line. Find the ratio of AC to CB. 

10. Show that if the coordinates of the vertices taken in 
order of a quadrilateral are (x v y 1 ), (x.,, y.^), (x v y.^j and 
( x v 2/4). its area is 

\ { x i (y 2 - vd + x 2 (y s - vi) + *» (y* - vd + - r 4 Oa - ^}- 

^14. In the A OAB, P is taken in OA, Q in AB and R 
in BO so that OP : PA = AQ : QB = BR : RO = 3 : 1. 
Show that A PQR : A OAB = 7 : 16. 



18 



ELEMENTARY ANALYTICAL GEOMETRY 



Loci 

20. The definition of a locus (see Ontario H. S. 
Geometry, page 77) is : — 

When a figure consisting of a line or lines con- 
tains all the points that satisfy a given condition, 
and no others, this figure is called the locus of 
these points. 

The condition which the points satisfy may be 
expressed in the form of an equation involving the 
coordinates of the points. For example, take the locus 
of the points of which the ordinate is equal to 3. 
This condition, which is expressed by the equation 
y = 3 [i.e.: — Ox + y = 3], is satisfied by an infinite 
number of points, as (0, 3), (1, 3), (2, 3), (7, 3), (-4, 3), 
etc. All such points are on a st. line AB || to Ox and 3 

y 



■8 



Fig. 13. 



units above it ; and this st. line contains no points which 
do not satisfy the condition. Thus the equation y = 3 
represents the line AB. 



LOCI 



19 



Similarly the equation y = — 3 represents a st. line 
|| Ox and three units below it ; x = 3 represents a st. 
line || Oy and three units to the right of the origin, 
and x — — 5 a st. line || Oy and 5 units to the left 
of the origin. 

For another example let us take the condition to be 
that the abscissa and ordinate of each point are equal. 
The points (0, 0), (1, 1), (2, 2), (4, 4), (-.1, -1), 
( — 5, —5), etc., satisfy this condition. It is expressed 
by the equation y = x. If we draw a diagram on 



H- X 


_l2,tk? _ 


H * T 5?^AT 


7 


^ 


t*.ikZ 




7 




2 


-* y 


~" y' 


7 


. 7 


ZH^r 


7 


7 


TC7 


23-2^ ^ r 





Fio. 14. (Unit = A inch.) 

squared paper, mark some of these points on it and 
join them we get a st. line AB bisecting the Ls xOy 
and x'Oy' every point on which satisfies the given con- 
dition. Between O and (1, 1) there are an infinite 
number of points, (h, h), (J, £), ( T V, T V), ( T V, T V), etc., 
which satisfy the condition, and so on continuously 
throughout the line. Thus the equation y = x represents 
the line AB. 



20 



ELEMENTARY ANALYTICAL GEOMETRY 



Again, we may consider the point which moves so 
that its distance from the origin is always 5. Its locus 
is plainly the circumference of a circle. Particular 



1 1 


" (0.5) 


J-3.4)y^ ^S,d 3 ^) 




/- V( 4 - 3 )- 




/ s 


~7 C 


f \ 


' 1 'I 


t-5.0) \S&) ' 


*._ _£ _ 


: t _l jt 


x t 


X t 


\ ^ t 


-L V -J 


\ -Z 


± 5 z 


-^s- — io-sv^- 


: z *-Jt ::: :::: 



Fig. 15. (Unit = ^ inch.) 

points on this locus are (5, 0), (4, 3), (3, 4), (0, 5), 
( — 3, 4), etc., and its equation is Jx 2 + 2/ 2 = 5, or 



+ 2/ 2 



25. 



21. In the equation of a locus the numbers that 
are the same for all points on the locus are called 
constants ; while those that change in value con- 
tinuously from point to point are called variables. 

Thus, in the equation x 1 + y 2 = 25, x and y are 
variables and 25 is a constant. 



EXERCISES 21 

22.— Exercises 

sJl. Find four or five points on the iOcus represented by 
each of the following equations ; and draw the locus on 
squared paper in each case : — 

(a) x = - 4 ; (6) x + y = ; (c) x - 2y = ; 

ifi 3x + y = j (e) x = y+i; (/) ^ + ^ = 169. 

v ^2^ A point moves so that its distance from the axis of 
a: is 5 times its distance from the axis of y. Find the 
equation of its locus. 

v Jk What locus is represented by the equation (a) y = ; 
(6) x = 0? 

M^A point moves so that it is equidistant from the 
origin and from (8, 0). Find the equation of its locus. 

v/jx A point moves so that it is equidistant from the 
origin and from (3, - 5). Find the equation of its locus, 
and draw the locus on squared paper. 

v<6r A point is equidistant from (1, -2) and ( — 3, -4). 
Find the equation and draw the locus on squared paper. 

7. A point moves so that its distance from (4, 3) is 
always 5. Find the equation and show that the locus 
passes through the origin. 

*- 8. The coordinates of the ends of the base of a A are 
(-2, -3) and (4, -1), and the length of the median 
drawn to the base is 6. Find the equation of the locus of 
its vertex. 

9. The coordinates of the ends of the base of a A are 
(0, 0) and (5, 0), and its area is 10. Show that the 
equation of the locus of its vertex is y = 4. 

V10. The coordinates of the ends of the base of a A are 
( - 1, -2) and (5, 1) and its area is 9. Find the equation 
of the locus of its vertex. 



22 ELEMENTARY ANALYTICAL GEOMETRY 

23. An equation connecting two variables x and y 
has an infinite number of solutions. For example, in 
the equation y = 3 x + 7, if any value is given to x, 
the corresponding value of y may then be determined. 
Thus, when 



(a) x = 


0,2/ = 


7, 


(b)x = 


1,2/ = 


10, 


(c) X = 


2,2/ = 


13, 


(d)x = 


-1,2/ = 


4, 


(e) x = 


-3,2/ = 


-2, 


(f)x = 


hy = 


8, 



etc. 

The, in general, continuous line which passes through 
all the points (a), (6), (c), etc., is the locus represented by 
this equation. 

Another equation as 4>x + 3y = 8 has also an infinite 
number of solutions, and if these two equations are 
solved together, the common solution obtained, in this 
case x = —1, y = i, gives the coordinates of the point 
of intersection of the loci represented by the equations. 

Sets of solutions which satisfy the equation 
4cc + Sy ;= 8 are given in the following table :— "■ 



(cZ)-l 

{S) 2 

(A) 5 



y 



EXERCISES 23 

If we plot these two sets of results on squared 




Fig. 16. (Unit = & Inch.) 

paper, we see that the loci appear to be st. lines 
which intersect at the point (d) ( — 1, 4). 

24.— Exercises 

X. Plot the following loci on squared paper and find the 
coordinates of their points of intersection : — 

tyrf 4a; - y >= 1 and x - 2y = - 12 

(b) x + 2y = 7 and 5x - 2y = 11 

(c) 3,r + 8y = - 18 and 4r + 3y = - 1 
(cl) 3.c -f 4y = and x 2 + y 2 = 100 
(e) 3a: - 5y -f 45 = and x 2 + y 2 = 169. 

V2f Find the points where the locus 3x - 5y -f- 45 = 
cuts the axes. 

3. Find the points where the locus x 2 + y' 2 = 6 x cuts 
the axis of x. 



24 ELEMENTARY ANALYTICAL GEOMETRY 

4. Find the locus of a points such that the square of its 
distance from ( - a, o) is greater than the square of its 
distance from (a, 6) by 2 or. 

•5. Find the equation of the locus of a point such that 
the square of its distance from (-2, - 1) is greater than 
the square of its distance from (5, 3) by 11. 

6. A (1, 0) and B (9, 0) are two fixed points and P is a 
variable point such that PB = 3 PA. Find the equation of 
the locus of P. 

7. Plot the following loci and show that they are 
concurrent : — 

%x + 4y = 10, 5x - 2y = 8, 4x -f y - 9. 



CHAPTER II 

The Straight Line 

(25. To find the equation of a st. line in terms of 
the intercepts that it makes on the axes. 




Let the st. line cut the axes at A, B so that OA 
OB = b. 



a, 



Take P (x, y) any point on the line, and draw PM \\Oy 
and terminated in Ox at M. 

From the similar As APM, ABO, 



PM AM 
BO ~ AO " 






y a—x, 
b a 






a + b L 







Note. — It is seen from the diagrams that both the proof 
and the form of the equation are the same for oblique and 
rectangidar axes. 

25 



26 



ELEMENTARY ANALYTICAL GEOMETRY 



2G. To find the equation of the st line passing 
through A (x v y x ) and B (a^ y 2 ). 

P. 




Take any point P (x, y) on the st. line. 

Draw AK, BL PM || Oy and terminated in Ox at 
K, L, M ; and AN, BR || Ox and respectively terminated 
in PM at N and AK at R. 

From the similar As PNA, ARB, 



AN 
BR 



PN 
AR' 



AN = KM = OM - OK = X - X v 
BR = LK = OK - OL = x 1 - X 2 , 
PN = PM - NM = PM - AK = y - y v 
AR - AK - RK = AK - BL = y x - y 2 

. x - x t _ y - y t 
x L -x 2 y!-y 2 ' 

Note. — It is seen from, the diagrams that both the proof 
and the form of the equation are the same for oblique and 
rectangular axes. 



EXERCISES 27 

27. — Exercises 

1. The equation of the st. line passing through (4, 3) and 
(-2, 7) is by the formula of § 26 

x - 4 y - 3 

4+2 = 3^7 <* 

or, 2x + 3y = 17. 

To find the intercepts which this line makes on the axes, 
let y = and .*. x = 8£, let a; = and .*. y = 5f. By § 25 
the equation of the line may now be written 

8i + 5f 
This is clearly the same as 2x -f Zy = 17. 

2. Write down the equations of the st. lines which make 
the following intercepts on Ox, Oy respectively : — 

(a) 5, 2; (b) -4, -6; (c) 3, -8. 

3. Find the equations of the st. lines through the follow- 
ing pairs of points : — 

{a) (6, 2), (3, 1); (b) ( - 1, 2), (-3, -7); (c) (4, -6), 
( - 7, 2). Find the intercepts these st. lines make on the 
axes. 

4. Find the point where the st. line which makes inter- 
cepts - 3 and 5 on Ox and Oy respectively is cut by the 
st. line x = - 5. 

">. Find the point where the st. line making intercepts 7 
and 2 on Ox and Oy respectively meets the st. line through 
( - 2, 7) and (5, - 3). 

6. Find the point where the st. line through (3, 5) and 
(-7, - 1) meets the st. line through (-8, 2) and (6, 5). 



28 ELEMENTARY ANALYTICAL GEOMETRY 

1 7. Prove that (11, 4) lies on the st. line joining (3, -2) 
and (19, 10) and find the ratio of the segments into which 
the first point divides the join of the other two. 

8. Find the equations of the sides of the A of which 
the vertices are (4, -2), (-5, -1), and (-2, -6). Find 
also the equations of the medians of the A and the coordi- 
nates of its centroid. 

9. The vertices of a quadrilateral are (3, 6), ( - 2, 4), 
(2, - 2) and (7, 3). Find the equations of the four sides. 
Find also the equations of the three diagonals of the 
complete quadrilateral, and show that the middle points of 
the diagonals are collinear. Find the equation of the st. 
line passing through the middle points of the diagonals. 

10. Find the vertices of the A the sides of which are 
11.x - 3y = - 45, 5x - lly = 47 and 3x + ly = 7. 

11. P (ajj, y x ) is any point and - + ''- = 1 cuts Ox, Oy at 
A, B respectively. Show that the area of the A PAB = 

l (bx x + ay x - ab). 



THE STRAIGHT LINE 29 

28. As explained in elementary algebra, the degree 
of a term, with respect to certain letters, is the number 
of such letters that occur as factors in the term. 

Sx, — by, ax, by are terms of the first degree with 
respect to x and y. 

5x 2 , Sy 2 , —Ixy, ax 1 are terms of the second degree 
with respect to x and y. 

29. Degree of an equation. An equation is said to 
be of the first degree in x and y when it contains a 
term, or terms, of the first degree in x and y, but no 
term of a higher degree than the first. 

The general equation of the first degree in x and 

V is 

Ax + By + C = 0. 

An equation is said to be of the second degree in 
x and y when it contains a term, or terms, of the 
second degree in x and y, but no term of a higher 
degree than the second. 

The general equation of the second degree in x 

and y is 

Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0, 
or, in a more convenient form, 

ax 2 + 2hxy + by 2 + 2gx + 2fy -f c = 0. 



30 ELEMENTARY ANALYTICAL GEOMETRY 

30. To prove that an equation of the first degree 
always represents a st. line. 

Let (x lt 2/1), (#2> 2/>)> ( x z> 2/3) De an y three sets of 
simultaneous values of x and y which satisfy the 
equation Ax -\- By + C = 0. 

Then, (1) Ax- X + By 1 -j- C = 0. 

(2) Ax 2 +By 2 + C = 0. 

(3) Ax s + B7/ 3 + C = 0. 

From (1) and (2), 

A B C 



2/l - V-2 X 2 ~ X l X lV 2 - X 2Vl 

Dividing the three terms of (3) respectively by 
these equal fractions and by any one of them, 

x s (Vi ~ 2/2) + 2/3 O2 ~ x i) + x i V-2 - » 2 Vx = °- 
Rearranging the terms, we get 

( 4 ) x i (2/2 ~ 2/3) + ® 2 (2/3 ~ 2/i) + »3 (2/i - 2/2) = 0- 
From § 17 the area of the A formed by joining 
(x v y x ), (x 2 , y 2 ), (x 3 , 2/3) is 

I { x i (2/2 ~ 2/3) + X 2 (2/3 ~ 2/i) + *3 (2/i ~ 2/2)} 
and .'., from (4), in this case, the area of the A is 
zero. 

This can only be so when the three points are in a 
st. line, and .'. as any three points the coordinates of 
which satisfy 

Ax + By f C = 

are in a st. line, this equation must always represent 
a at. line. 



THE STRAIGHT LINE 31 

(3Lj The equation Ax -{-By -f C = may be changed 
to the form 

-A_ + -£-=!, 
C _ C 
A B 

and by comparing this with the equation of § 25, 

x y 

a b 
we see that the intercepts which the st. line 
Ax + By + C = makes on the axes of x and y are 
respectively 

c , c 

- a and - g. 

The same results are obtained by alternately letting 
y = and x = in Ace + B^ + C = 0. 

32. To obtain the result of § 26 from the general 
equation of the first degree. 

Let (x v y x ), (x 2 , y 2 ) be fixed points on the st. line 
represented by the general equation, and we have 

(1) Ax + By + C = 0, 

(2) Aa^ + By 1 + C = 0, 

(3) A.r 2 + By 2 + C = 0. 
From (2) and (3), 

(4) _A_ = — B — = 5 . 

2/1 — 2/2 as, - a* a^ 2 - a^ 
.*. , from (1) and (4), 

as (2/i - 2/2) + 2/ O2 - aO + x x y. 2 - x 2 y x = 0. 

This equation is seen to be the same as 

x - x, = y - y x 

asi - ^2 2/i ~ 2/2' 

when the latter is cleared of fractions and simplified. 



32 



ELEMENTARY ANALYTICAL GEOMETRY 



3.3. To find the equation of a st. line in terms of 
its inclination to the axis of x and its intercept on 
the axis of y. 



y 


B 








N 















M 


* — ^ 


v * 



Fio. 21. 

Let the st. line cut Ox, Oy at A, B respectively, 
L BAx = a, and OB = b. 

Take any point P (x, y) in the line, and draw PM 1 

Ox, PN _L Oy. 

m r,™ BN BO - PM h - y 

Tan BPN - ^ = om = -5— 

But, tan BPN = tan PAM = - tan a. 

b - y 



tan a 



x 



and :. y = x tan a + b. 
If we let tan a = m, the equation becomes 
y = mx + b. 



THE STRAIGHT LINE 33 

In this equation m is called the slope of the line, 
and the Z a, or tan~ l m, is always measured by a 
rotation in the positive direction from the positive 
direction of Ox, i.e., the Z is traced out by a radius 
vector starting from the position Ax and rotating 
about A in the positive direction to the position AB. 

Note. — For oblique axes the proof and result are different 
from those given above for rectangular axes. 

(3C) The equation Ax + By + C = may be changed to 
A C 

y = - b x - b' 

from which by comparison with 
y = tux + b, 
it is seen that the slope of the st. line Ax + By -{- C = § 



_ =» and its intercept on the axis of 



is — 




B 



3 - - jcj) ^yfmjjL 



34 



ELEMENTARY ANALYTICAL GEOMETRY 



35. To find the equation of a st line in terms of 
the j_ on it from the origin and the l made by a 
positive rotation from Ox to this _l_. 





y 




o 


N ^S^ 


X 









Fio. 22. 

Let the jl OM from O to the line = p, and 
I #OM = a. 

Take any point P (x, y) in the st. line. 

Draw PN j. Ox, NRi OM, MH I! Oy to meet NR at H. 

OR + RM = p. 

OR = ON cos RON = x cosa. 

RM == MH cos RMH = PN cos MOy = y sum. 

.'. x cos a + y sin a = p. 

u Note. — i^or oblique axes the proof and result are different 
from those given above for rectangular axes. 

36. To reduce the equation Ax + By + C = to 
the form x cos a + y sin a = p, where p is always a 
positive quantity. 

The equations 

x cos a + y sin a — p = 
Ax + By + C =0 

will be identical if 



cos a sin a — p 

- f / tit— 

* 




-*B 2 



THE STRAIGHT LINE 
If C is a positive quantity, 



p _cos a _sin a _ y ' cos 2 « + sin 2 « 
C _ ~^~A ~^TT ~ 



t/a 2 4- B 2 y A 2 + B 2 

- A - B _ C 

.'. cos a = ■ , ' , sin a = — , and p = 



y'A* + B 2 i/A 2 + B 2 V A 2 + B 2 

If C is a negative quantity, these results should be 
written 

A B -C 



cos a ==r , sin a _ , p 

- / A 2 I Di -./A2I D 2 ■*• 



y A 2 + B 2 ' i / A 2 4- B 2 '^ /A 2 + B 2 



Thus the equation is 



AX =pBy = _±C 



V/A 2 + B 2 v'A 2 + B 2 /A 2 + B 2 ' 



\ 



the upper signs being taken when C represents a 
positive quantity and the lower signs when c repre- 
sents a negative quantity. 

37. Ex. 1. Reduce the equation Sx + 4y — 12 = to 
the form x cos a + y sin a = p. 

Here j/j^T"! 2 = l/25 = 5. 
Dividing the given equation by 5 
3 4 12 

cos a — -, sin a = , and .*. a = tan ( \ while the 

12 

X from the origin on the line is — 

Ex. 2. Reduce the equation x — y -\- 7 = to the 
form x cos a + y sin a = p. 

t" fr^><- ± "&X- 

, "~ — » ~i — r~r= — : 



* *B X 



36 



ELEMENTARY ANALYTICAL GEOMETRY 



Here |/P + 1' = V 2. 

Dividing the given equation by — y/2 









05 1/ 

i/2 i/2 


7 
~i/2 








or, 


35 COS 135° 


+ y 


sm 135° 


7 
~ i/2 


i.e 


, a = 


135° 


, and the 


J_ from the 


origin 


line 


. 7 
is — - 
i/2 













38. To find the equation of a st. line in terms of 
the coordinates of a fixed point on the line and 
the z which the line makes with Ox. 




Let Q (x v y{) be the fixed point and 6 the Z. 

Take any point P (x, y) on the line and let QP = r. 
Draw PM, QN j_ Ox and QR ± PM. 
qr = pq cos PQR. 
QR = NM = x — x u and Z PQR = Z Q. 

X — x x 

it = r - 

cos 6 



PR = 


PQ 


sin 


PQR. 




PR = 


PM 


- RM = 


PM 


2/ - 
sin 


2/i 

e ' 


= r. 






x - 


*l 


_ y 


-yi 


= ; 



THE STRAIGHT LINE 37 



QN = y - y v 



cos 6 sin 

This form will frequently be found useful in pro- 
blems that involve the distance between two points on 
a st. line. 

If cos 6 = 1 and sin 6 = m, the equation becomes: — 

*-* i _ y-yi _ r 

1 m 

I and m are called the direction cosines of the st. 
line, and I 2 + m 2 = i. 

39. For convenience of reference the different forms 
of the equation of the st. line are here collected : — 
(1) Ax + By + C = 0. 







(2)- + 

a 


y 
b 


= 1. 










(3) X " 


x x 


_ 2/ ~ 2/i 
2/i - 2/2 

+ y sin a 






C8) *! - 

(4)2/ = 
(5) x cot 


x 2 

nt.r 
1 a 


= 7?. 


(6) " - 

If the st. line passes 
e convenient form : — 


x i - y - 2/1 _ 

m 
through the > 


= r. 

arigin (4) takes 






0)y = 


in,' 









38 ELEMENTARY ANALYTICAL GEOMETRY 

If the st. line joins the origin to a fixed point 
(Ph> Vi)> we get, by letting x 2 = y 2 = in (3): — 

(8) X - = V - 

If the st. line passes through (x u y x ) and its slope is 
m, the equation is easily seen from (4) to be 

(9) y-y 1 = rrb{x -x x ). 

40.— Exercises 

1. Name the constants and variables in each of the nine 
equations of § 39. Explain the meaning of each constant. 
Which of these equations are of the same form for rectangular 
and oblique axes 1 

; '2. , Draw the following st. lines on squared paper : — 

(a) x + 2y = 8 ; (b) 3x - ly = - 5 ; (c) |- + | - - lj 

(d) 2x + 3y = 13; (e) Sy = 4x. 

3. Find the equation of the st. line. 

(a) through the origin and making an Z of 30° with Ox; 

(b) through the origin and making an Z of 120° with Ox ; 

(c) through (0, 5) and making the Z tan _1 f with Ox ; 

(d) through (0, -3) and making the / cos -1 £ with Ox; 

(e) through ( - 3, - 4) and making the Z. 45° with Ox. 

4. In the A of which the vertices are ( — 2, 5), (3, - 7), 
(4, 2) 

(a) find the slope of each side; 

(b) show that the medians are concurrent and find the 
centroid. 

5. O (0, 0), A (6, 0), B (4, 6), C (2, 8) are the vertices 
of a Quadrilateral. Show that the st. lines joining the 



THE STRAIGHT LINE 39 

middle points of OA, BC, of AB, CO and of OB, AC are 
concurrent, and find the coordinates of their common point. 

•fl^Find the equation to the st. line through ( - 4, 3) that 
cuts off equal intercepts from the axes. 

7. Find the length of the _L from the origin to the line 
Bx + 7y = 10; find the Z which this ± makes with Ox. 

v 8. What is the condition that the st. line Ax -f- By + C = 
may 
v (a) pass through the origin; 
■■(b) be || Ox ; 
J (c) be || Oy ; 

- (d) cut off equal intercepts from the axes ; 
v (e) make I 45° with Ox; 

/ 9T'~'\Vhat must be the value of m if the line y = mx -j- 7 
passes through ( - 2, 5) 1 

y 1-0.' Find the values of m and 6, if the st. line y = moo + 6 
passes through ( - 2, 3) and (7, 2). 

v [11. Find the values of a and b, if the st. line _ -f ^ = 1 

a b 
passes through ( - 2, - 5) and (4, - 2). 

JL2. Show that the points (4a, -36), (2a, 0), (0, 36) are 
in a st. line. 

\ 13. Show that the intercept made on the line x = k by 
the lines Ax -\- By + O = Q and Ax + By -f- C' = is the 
same for all values of k. 

14. P (x v y x ) is any point and the lines Ax -f- By + C = 
cuts Ox, Oy at N, R respectively. Show that /\ PNR = 

2^B (Ax i + B ^i + C >' 



40 



ELEMENTARY ANALYTICAL GEOMETRY 



The Angle Between Two Straight Lines 



41. To find the Z between two st. lines whose 
equations are given. 

(i) Let the given equations be y = m x x + 6 X and 
y = m.jX + b 2 . 




Let AB be the line y = m x x + b x and AC be the line 
y = m.jX -f b 2 when B, C are on the axis of x. Let 
L BAC = 6. 

Then n\ = tan AB£, m 9 = tan AC#. 



.*. tan 6 



z 6 = z ABj? — z ACr 

fcm AB21 — ton AC.s 



-tc^G-rto^* 



_ HC*- 



w. 



= tan 



1 + tan AB£ . tan ACx 1 + m 1 m 2 
m, — m., 



1 + m^ 

(fn)) Let the given equations be Ax + By + c = and 
A,a; + B,2/ + Cj = 0. 

These equations may be changed to 

„=-**-§ and y = -£*-| 



THE ANGLE BETWEEN TWO STRAIGHT LINES 41 

.*. writing — — for m, and — — 1 for m 9 in the above 
result, the Z between the lines 
_ ^ + A x 

+ BBj 

(^42) Condition of Parallelism. If two st. lines are 
||, they make equal Zs with the axis of x, i.e., their 
slopes are the same. 

.*. , if their equations are y = m r « + b x and y = 
rn<£C -f 6 2 , the condition is 

m 1 = m 2 . 

If their equations are Ax + By + C = and 
A x x + B^ + C x = 0, the condition is 
_ A _ _ A, 
~ B ~ b/ 

or, AB X - AiB = 0. 
A B 

This may also be written t = o ', and we see that 

the equation ax + by = k can be made to represent 
an infinite number of || st. lines by giving different 
values to k ; as : — ax + by = k x , ax + by = k 2 , etc. 

(J3. Condition of Perpendicularity. If the st. lines 

y = m x x -f- b x , y = m.pc + b 2 are j_, 

tow _1 — ! ?_ _ «_^ 

1 + mjWig 2 

1 ■+■ m x m<, 



42 ELEMENTARY ANALYTICAL GEOMETRY 

This will be true if 1 + m x m 2 = 0, and .-. the 
required condition is 

n^m* = - 1. 

Similarly, if Ax + By + C = and A x x + B x y + C L = 
are J_. 

AA X + BB : = 0. 
The st. lines 

Ax + By + C = 

Bx - Ay + Cj= 
satisfy the above condition and .*. are ± to each 
other. 

44.— Exercises 

1C Find the L between the st. lines 
-(a) 2x - 3y = 9 and x + 5 y = 11 ; 
(6) 3x + 5y = 12 and (17/3 + 30) x + 33y = 19; 
■ (c) by = 3x + 12 and 5cc + 3y = 17; 
(J) ix + ly = 13 and 3* -?/ = 6. 

2. Find the equation of the st. line || to 6a? - ly = 13 
and passing through ( - 2, - 5). 

Solution: — The required equation is 

6 (x + 2) - 7 (y + 5) = ; 
i.e., 6x - ly = 23. 

3. Find the equation of a st. line through ( - 3, - 5) 
and || to 9x + 4y = 18. 

4. Find the equation of the st. line drawn through ( - 2, 
- 5) and J_ to 6a; - ly =13. 

Solution: — The required equation is 

7 (x + 2) + 6 (y + 5) = ; 
i.e., lx + 6y + 44 = 0. 

5. Find the equation of the st. line drawn through (4, 2) 
and J_ 3x -2y = I. 



EXERCISES 



43 



^6. Find the equation of the st. line passing through 
(-3, 5) and || to the st. line joining (2, 6) and (7, -1). 
^m Find the equation of the st. line passing through (2, 6) 
and JL the st. line joining ( - 3, 5) and (7, - 1). 

V8f Find the equations of st. lines drawn through (5, 7) 
which make Zs 45° and 135° with Ox. 

\>9. Find the equations of st. lines drawn through ( - 5, - 3) 
which make ^s 30° and 150° with Ox. 

v^lCJ. Show that the _l_s from the vertices of the A (1, - 3), 
( - 5, - 2), (4, 7) to the opposite sides are concurrent ; and 
find the coordinates of the orthocentre. 

11. Show that the J_s from the vertices of the A (0, 0), 
(a, 0), (b, c) to the opposite sides are concurrent ; and find 
the orthocentre. 

12. Find the ratio into which the _L from the origin on 
the st. line joining (2, 6) and (5, 1) divides the distance 
between these points. 

03.;Find the equations of the st. lines which pass through 
(h, k) and form with y = mx + b an isosceles A of which 
the vertex is at the given point and each base Z = a. 




n, 



Solution : — Let y - k = M (x - h) represent one side of the A 
where the value of M is to be found. 



44) ELEMENTARY ANALYTICAL GEOMETRY 

Tan a ■■ 



M 



m M 



m + tan n 



1 - m tan a 

.'. the equation of this side is 

7 m + tan a , ■,, 

y - h= . (x - h). 

1 - m tan a 

If for a we substitute tQ8° - a, the equation of the other side is 
found to be 

y-k= » - *» * (X - A). 

1 + m <om a 

^k Find the equations of the st. lines passing through 
(2, 8) and making an I of 30° with 3x - I2y = 7. 

^l&T Find the equations of the st. lines passing through 

(-1, -2) and making an Z of 45° with * + | = 1. 

16. Show that the equation of the st. line through (a, b) 
and making an Z of 60° with x cos a + y sin a = p is 
y — b = {x - a) tan (u ± 30°). 

17. Show that the right bisectors of the sides of the A 
(0, 0), (a, 0), (b, c) are concurrent ; and find their point of 
intersection. 

18. Find the equation of a st. line J_ to Ax + By f C 
= 0, and at a distance p from the origin. 



PERPENDICULARS 



1,5 



Perpendiculars 

(45) To find the length of the _L from P (x lt y x ) to 

A^ + Bi/ f C = 0. 

y 




Draw PM _L the given st. line. Join P to N, R 
the points where the given st. line cuts Ox, Oy. 
A PRN =1PM. RN. 



ON = — ^ and OR = - 



RN 



B 2 . 



By the formula of § 17 

A 



= - 2Xi (A ri + B ^ + c >- 



•'• i PM • ?b\/ A2+ B ' = " 2a¥ (a *> + B ^ + c > 
A^ + By, + c 



;. PM = - 



The length of the _]_ is .'. 

ax x + By x + C 
•a» 4- B2 



/A*+B» 



46 



ELEMENTARY ANALYTICAL GEOMETRY 



In the diagram AB is the line represented by the 
equation Ax + 5y — 20 = 0. 



, . 






1 


si; y - 








% 
















->«h ^ s 








*V 






-5^>i- 


N 


v 








N ^ 








N> 




.. 




S 


Tj 






V 
















- ^ -, 








V 5 J 








"v 











If 



- 20. 

- 25. 

- 27. 
+ 30. 
+ 35. 



Fio. 27. (Unit = T '„ inch.) 

the expression Ax + 5y - 20 we substitute 
the coordinates of points O, P, Q, R, S which are not 
in the line, the following results are obtained. 
For o (0, 0), Ax + 5y - 20 = 
-- P (5, -5), 4x + 5y - 20 = 
m Q (-8, 5), 4cc + 5y - 20 = 
.. R (5, 6), Ax + 5y - 20 = 
ii S (10, 3), Ax + 5y - 20 = 
In these results it will be observed that: — 

For the origin the sign of the value of the expression 
is the same as the sign of the absolute term. 

For other points that lie on the same side of the 
given st. line as the origin the signs of the values of 
the expression are the same as the sign of the result 
for the origin; while, for points on the side remote 
from the origin the signs of the values of the expression . 
are different from the sign of the result for the origin. 



PERPENDICULARS 



47 



A formal proof of these properties is given in the 
next article. 

46. To prove that the sign of the expression A* 
-f- By + c is different for points on opposite sides 
of the line ax + By + G = 0. 





y 




3 
















Jk"^ 










N 




-Q 




^ f 





\ 


1 


R 


— ? 



p (^h' Vi)> Q ( x -2> Vd are an y points on opposite sides 
of Ax + By + C = 0. 

Draw PM, QR _L Ox and let them cut the given line 
at N, S. 



Vi 



PM 



PN + NM ; y, = QR = SR - SQ. 



.-. AX X + By x + C = AiCi + B.PN + B.NM + C, 
and Ax 2 + By 2 + C = Ax. 2 + B.SR - B-SQ + C. 
But, V N and S are both on the given line, 
Ax x + B.NM + C = 0, 
and Ax, + B.SR + C = 0. 
.*. A£Ci + By x + C = B.PN, 
and Ax 2 + By 2 + C = — B.SQ. 

.*. , since PN, SQ are both taken as positive quanti- 
ties, Ax x + By x -f C and Ax 2 + By 2 + C have opposite 
signs. 



48 



ELEMENTARY ANALYTICAL GEOMETRY 



When x = and y = the expression Ax + By 4- C 
becomes C, .". a point whose coordinates when substi- 
tuted in Ax + By + C gives the same sign as C 
is on the same side of the st. line Ax + By + C = 
as the origin. 

47. Sign of the Perpendicular. It follows from 
the preceding article that, if the positive sign is 
always taken for -j/A" 2 + B-, when the sign of 
A£C, + B//, + C 



/A2 4- B2 
is the same as the sign of C, the point (x v y x ) and 
the origin are on the same side of the line Ax + By 
+ C = 0; and when the sign of this fraction is 
different from that of C, the point (x v y x ) and the 
origin are on opposite sides of Ac + By + C = 0. 

o find the equations of the bisectors of the 




zs between the lines Ax 4- b</ 4- c = and A x jr 4- 
Btf 4- C x = 0. 



PERPENDICULARS 49 

The Is to the st. lines from any point P (x, y) on 
either bisector are equal to each other, 

.*. the required equations are 

Ax + By + c = ^ Aj x+_B 1 y_+_c 1 

^A 2 + B 2 V A x 2 + Bf 

If the equations are so written that C and C x have 
the same sign and P, is on the bisector of the L that 
contains the origin, the JLs from P, have the same 
sign as the J_s from the origin on the lines, and the 
equation of the bisector is 

Ax + By + C A x x + B l2 / + C a 



V A 2 + B 2 VAs + Bj 2 

If P is on the bisector of the i_ which does not 
contain the origin, the Is from P have opposite signs 
and the equation of the bisector is 

Ax + By + C A X X + B x y + C x 

l^A 2 + B 2 



50 



ELEMENTARY ANALYTICAL GEOMETRY 



(49.^ To find the distance from (a, b) to Ax + By + c = 
in the direction whose direction cosines are /, m. 

The equation of the st. line passing through (a, b) 
in the given direction is, by § 38, 

x — a y — b 

I 711 

:. x = a + lr, y = b + mr. 

Substituting these values for x and y in Ax -\- By + 
C = 0, 

A<x + Air + B6 + Bmr + C = 0. 

_ Aa + Bb + c 
Al + Bm 

50. The length of the J_ from (a, b) to Ax + By + 
C — may be deduced from the result of § 49. 

a y - b 



For, if 



and Ax + By + C = are 



_]_ to each other, 
A _ B 
I m 

since I 2 + m 2 = 1. 



A£ + Bm 



P + m 2 
Also each of these fractions 



= Al + Bm, 



VA 2 + B 2 



l/A 5 



B 2 . 



VI 2 + m l 
:. Al + Bm = VA 1 + B 2 , 
and the length of the ± is 

Aa + Bb + C 
i/A 2 + B 2 




PERPENDICULARS 51 

51. To find the equation of a line passing through 
the intersection of two loci. 

The equation 

k (ax + By + C) + I (A x « + B lV + CO = 0, (1) 
being of the first degree in x and y represents a st. line. 
If (x v 2/ x ) is the point of intersection of 

Ax + By + C = (2) 

and A^ + B x y + C 1 = 0, (3) 

the values x v y x substituted for x, y will plainly 
satisfy equation (1), and .*. the st. line (1) must 
pass through the point of intersection of the st. lines 
(2) and (3). 

From the same reasoning the following more general 
theorem is seen to be true: — 

If two equations are multiplied by any numbers 
and the results either added or subtracted, the re- 
sulting equation represents a locus that passes 
through the point (or points) of intersection of the 
loci represented by the first two. 

52. Example — Find the equation of the st. line passing through 
the intersections of 17x - ly = 9, 3x + 19y = 34 and J. to llx - 
4y = 13. 

17x - ly - 9 + I (3x + 19y - 34) = 
is a st. line passing through the intersection of the first two lines. 
This equation may be written 

(31 + 17) x + (19* - 7) y - 341 - 9 = 0. 
If this line is _1_ to llx - 4y = 13, 

11 (3/+ 17) - 4 (19/ - 7) = 
.-. 1 = 5, 
and the required equation is found to be 
32x + 88y = 179. 



52 ELEMENTARY ANALYTICAL GEOMETRY 

53. To find the condition that the three st. lines 

a x x + b x y + c x = (1) 

a 2 x + b 2 y + c 2 = (2) 

a s x + b 3 y + c, = (3) 

may be concurrent. 

If the three st. lines are concurrent, the coordinates 
of the common point satisfy the three equations. 

For that point, from (1) and (2), 

s = y = 1 

b x c 2 — b 2 c x c x a 2 — c 2 a x a x b 2 — a 2 b x 

Dividing the terms of (3) respectively by these 
equal fractions, 

a 3 (6 X c, — b 2 c x ) -|- & 3 (c x a 2 — c 2 a x ) + c. d (a x b.> — a 2 6 X ) = 0. 

This is the relationship that must hold among the 
constants in order that the lines may be concurrent. 



EXERCISES 53 

54. Exercises 

?{. Find the length of the _|_ 

(a) from ( - 4, 7) to 5x - 1y = 4 j 

(ft) from (- 4, -3) to| + | = 1 ; 

(c) from (3, - 2) to y = 7x + 1 ; 
"(c?) from ( - 2, - 7) to the st. line joining (5, 3) and 

(-3, -7); 
"(e) from the origin to the st. line joining (7, 0) and 

(0, -5). 

/2. Find the distance between the [| lines ix - 3y = 9, 
Ax -Zy = 2. 

x 3. Find the distance between the || lines ax -f by + c x 



^4. Find the point in the line — -f ^ = - 1 such that its 
J_ distance from the st. line joining (2, 7), (5, 3) is 8. 

* 5. Find the equation of the st. line through the intersec- 
tion of 3x - 2y = 12, 5x + iy = 9 and || to | + V - = 1. 
(See §§51 and 52). 

^6". Find the equation of the st. line joining the origin to 

the intersection of — f- f = 1 an( l T + - = 1. 
a b o a 

lZ?r Find the distance from the point of intersection of 
7x - by = 13, 4ic + 9y = 43 to the line 12a; = by. 

J^T'Find the equation of the st line passing through the 
intersection of y = mx -\- c, ?/ = n^a; -+- c x and J_ to 
a; y 

7, + f = L 



54 ELEMENTARY ANALYTICAL GEOMETRY 

ytf. Show that the st. lines x -J- 2y = 5, 2x + 3y = 8, 
3x -f j/ ■=> 5, x + y = 3 and 2a; - y = are concurrent. 

10. Find the condition that the lines ax -\- hy -\- g = 0, 
hx -{■ by -\- / = 0, ya: + /y -f c = are concurrent. 

lif Find the equation of the st. line passing through the 
intersection of y = mx -\- c, y = m-^x + C\ an d also through 
(a, b). 

\12. Find the equation of the st. line joining the origin 
to the point of intersection of Ax -f By + C = and 
* x x + B x y + Cj = 0. 

"13. Find the distance from the orthocentre of the A 
O (0, 0), A (8, 0), B (3, 5) to the st. line AB. 

14. Plot the lines 2x - 3y = 1, 3a; -f y = 7 on squared 
paper and find the intercepts that the bisectors of the Ls 
between them make on the axis of y. 

"' 15. Find the equations of the bisectors of the ^s between 
5x - 12y = 17 and 8a; + 15y = 31. 

16. Show that the bisectors of the supplementary Zs 
between y = mx -J- « and y = m r x- -f- a x are _|_ to each 
other. 

17. Find the equations of the bisectors of the L& between 
the st. lines joining (4, 5) and ( - 5, 2) respectively to 
(3, -7). 

v/18.' Show that the st. lines x -f 6y = 15, 2x - 5y -f 4 = 
and 9 a; + y = 29 are concurrent. 

19. The sides of a A are 3x + 4y = 15, 12a; - 5y = 17, 
24a; -j- 7y = 30. Plot the lines on squared paper and find 
the point where the bisectors of the interior Zs of the A 
intersect. 



EXERCISES 55 

,2©T Find the equation of the st. line passing through the 
intersection of the lines 2x - 3y — 6 and 3x + 4y = 18, 
and also through the middle point of the st. line joining 
(-1, 2) and (3, 4). 

i2?. Find the distance from (5, 3) in the direction in 

which the slope is — = to the line Ix - lly = 13. 
V3 J 

j^T Find the distance from (-4, 6) in the direction of 

which the slope is 1 to the line 1_ V- = \ 

r 2^3 

Draw the diagram on squared paper. 

-25T The sum of the distances from a point to the lines 
x + 2y = 7, 5x - 2y = 11 is 7. Show that the locus of the 
point is a st. line which makes equal Zs with the given st. 
lines. 

^JteT Find the distance between the || lines _ i £L _ c 

a o 

25. Find the equation of the st. line passing through P 
(2, 5) and cutting Ox at A, Oy at B so that AP:PB =7:3. 

26. Find the equations of the st. lines passing through 
(4, 7) and making an Z of 45° with 3x - 10 y = 8. 

27. Find the equations of the st lines passing through 
( - 4, - 7) and forming an equilateral A with 3x - 2y = 7. 

28. Find the equations of the st. lines drawn || to 
5x - I2y = 9 and at a distance 5 from it. 

29 Find the equations of the two st. lines which pass 
through (4, 7) and are equally distant from A (7, 3), 
B (3, - 1). Find also the distances from A and B to 
these lines. 



56 ELEMENTARY ANALYTICAL GEOMETRY 

30. Find the point in 4x - 3y = 12 which is equally 
distant from (2, 7) and (4, - 1). 

31. Having given the length of the base and the difference 
of the squares of the other two sides of a A, prove that the 
locus of its vertex is a st. line _L the base. 

32. Find the equations of the st. lines which are at a 
distance 2 from the origin and which pass through the 
intersection of a;- 7y-f 11 =0 and 3a? + 4y - 17 = 0. 

33. Find the equation of the st. line || to Ax + By -f 
C = and at a distance p from the origin. 

34. Find the equation of the st. line passing through 
(h, k) and _L to Ax + By -f C = 0. 

35. The equations of the sides of a A are 5x -{- 3y - 
15 = 0, 2x - y + 4 = 0, Sx - 1y - 21 = 0. (a) Show 
that the _l_s from the vertices to the opposite sides are 
concurrent and find the coordinates of the orthocentre. (b) 
Show that the right bisectors of the sides are concurrent 
and find the coordinates of the circumcentre. (c) Show 
that the centroid is at a point of trisection of the st. line 
joining the orthocentre to the circumcentre. 



CHAPTER m 

The Straight Line Continued. Transformation 
of Coordinates 

55. An equation of the second degree may represent 
two st. lines. 

For example, 2x 2 — 5xy + Sy 2 = is the same as 
(x —y) (2x — 3y) = 0, and will be true for all values 
of x and y which make either of the factors x — y or 
2x —3y equal to zero, and .". all points on the st. 
lines x — y = 0, 2x — Sy = are on the locus represented 
by 2x 2 - hxy + 3y 2 = 0. 

Similarly, an equation of the third degree may 
represent three st. lines, one of the fourth degree may 
represent four st. lines, etc. 

56. The general equation ax 2 + 2hxy + by 2 = 
represents two st. lines passing; through the origin. 

Solving as a quadratic in x 



h ± Vh- - ab 

— y> 



a 

from which it is seen that the given equation is 
equivalent to 



{ax + hy + y Vh 2 — ab} {ax + hy — y Vh 2 —ab) = 0, 
and .'. represents the two st. lines 



ax + hy + y Vh 2 — ab = 



ax + hy — y Vh 2 — ab = 0, 
both ot which pass through the origin. 



57 



58 ELEMENTARY ANALYTICAL GEOMETRY 

If h 2 > ab, the lines are both real. 
If h 2 = ab, the lines are coincident. 

If h 2 < ab, the lines are imaginary, and we have 
two imaginary sfc. lines passing through the real point, 
(0, 0). 

57. To find the z between the two st. lines 
represented by ax 2 + 2hxy + by 2 = 0. 

The given equation may be written 

2/H2^y + * x 2 = 0. 

If y — m x x and y — m^c are the factors of the 
expression on the left hand side of this equation. 

2h a 

m x + ra 2 = j- , m^z = ^-. 



.*. mj 2 -j- 2 m^g + m 2 2 = 



4A 2 
b 2 ' 



4 ra^j = 

.*. K - ™ 2 ) 2 = 62 


b ' 
ab) 


2 i//t 2 - 
and m, — m, = ; — 



then 6 is the Z between the st. 


ab 

lines, by §41. 


m, — m 2 2 Wi 2 — 

tort = — 3 L. = 

1 + m[m 2 o 


ab b 
X a + b 


2 Vh 2 - ab 





a + b 



THE STRAIGHT LINE CONTINUED 59 



...fl-toa- 8 ^'-^ 

a + b 

Condition of perpendicularity. If = 90°, 

tan 6 = oo. This will be the case if 
a + b = 0. 

58. To find the equation of the St. lines which 
bisect the zs between the st. lines represented by 
ax 2 + 2 hxy + by 2 = 0. 

Let the given equation represent the st. lines 

y — m r £ = 0, y — m 2 x = 0, so that 

2h a 

m x + m., = — __, m^m^ = r . 

o b 

The equations of the bisectors of the Zs between 
these lines are 

V - m i x , V ~ ™ >& y - m x x y - m,x 

z H , = and , — =0. 

vi+mf y\+m 2 2 ^l+mf Vi+ m2 2 

These equations may be combined into 

(y - m x xf _ (y - m 2 x) 2 = 
1 + m 2 1 + w 2 2 

Simplifying and dividing by ra 2 — m v 

(m x + w^) y 2 - 2 (mjm 2 - 1) xy — (m 1 + m 2 ) x 2 = 0. 

Substituting and multiplying by b. 

h (x 2 - y 2 ) - (a - b) xy = 0. 



60 ELEMENTARY ANALYTICAL GEOMETRY 

59. To find the relationship that must connect 
the constants in the equation 

ax" + 2 hxy + by- + 2 gx + 2fy + c =0 
in order that this equation may represent two st. 
lines. 

If the given equation represents two st. lines, it 
must be equivalent to two equations of the form 
y — m^x — b x = 0, y — m 2 x — b 2 = 9, from either of 
which y can be expressed in terms of the first degree 
of x. 

Solving the given equation for y, we obtain 



- (hx + f) ± V(hx + fY - b (ax 2 + 2 gx + c) 

y = - -5- — 

In order that these values of y may be in terms of 
the first degree of x, the expression under the radical 
sign must be a perfect square ; i.e., 

(/i 2 - 06) x 1 + 2 (hf - bg) x+f- be 
is a perfect square for all values of x. 

.-. (hf - bgf = (h* - ab) (f - be). 
Simplifying, we get the condition in the form 
2 fgh - ap - bg- - eh* + abc = 0. 

60. — Exercises 

1. Show that the following equations represent two st. 
lines and find the separate equations of the lines : — 

(a) x 2 - (a -f b) x = - ab ; (b) x 2 - y 2 = ; 
(c) x 2 - Zxy = ; (cl) 8x 2 f 3y 2 = 1 Oxy ; 
(e) xy + bx = ay + ab ; (/) 3x 2 - lOxy + 3y 2 - llx 
- 7y - 20 = 0. 



EXERCISES 61 

2. Show that 2x 2 - Ixy -f 6y 2 + 2s - 5y - 4 = repre- 
sents two st. lines and find the slope of each. 

3. Interpret the locus represented by xy = 0. 

4. Find the Zs between the st. lines in 1. (d), (e) and (/). 

5. Find the condition that axy + bx -f cy + d = may- 
represent two st. lines. 

6. Find the value of B for which the equation 3x 2 - 
lOxy + By 2 - 1x - 2y = 21 will represent two st. lines. 

7. Find the single equation which represents the two st. 
lines passing through (5, 3) and making an equilateral A 
with the axis of x. 

8. Prove that y 2 - Ixy sec a -f x- = represents two st. 
lines through the origin and inclined to each other at an 
L = a. Show also that one of these lines makes the same 
Z with the axis of x that the other makes with the axis 
of y 



ELEMENTARY ANALYTICAL GEOMETRY 



Transformation of Coordinates 

61. It is often necessary to change the coordinates 
involved in a problem into a different set which are 
referred to axes drawn 

(a) from a new origin, or 

(b) in directions different from the original axes. 

62: To change from a pair of axes to another 
pair which are || to the former, but have a different 
origin. 



V 


Y 


ij<) 


N 




Q 




X 





R 


r 


^ f 



Fig. 30. 

Q (h, /,•) is the new origin. 

Let P (x, y) be any point referred to Ox and Oy 
and X, Y the coordinates of the same point referred 
to the new axes QX and QY. 

Draw PNM ± to Ox and QX, and let YQ cut Ox 
at R. 

X = OM = OR + QN = h + X. 

y = PM = QR + PN = k + Y. 



TRANSFORMATION OF COORDINATES 



63 



Thus, if for x, y respectively we substitute h + X, 
k + Y in any equation the origin is changed to the 
point (h, k). 

To return to the original origin the substitutions 
would be X = x — h, Y = y — k. 

(^63^-To change the direction of the axes, without 
changing the origin, the axes being rectangular. 



y 


^^^.CC 


s 


I 

i 
i 


^y 




y/O 




I 


A R 




X 



Let P (x, y) be any point referred to Ox, Oy ; and 
X, Y the coordinates of the same point referred to 
axes OX, OY such that I XOa; = a. 

Draw PM J_ Ox, PN J_ OX, NR _|_ Ox, NS _L PM. 

Z NPS = 90° - Z NAP = 90* - Z MAO = a. 

x = OM = OR - NS = X cos a - Y sin a. j 

V = PM = NR + PS = X sin a + Y COS a. I 



T* 






Thus, if for x, y we substitute respectively X cos' a 
— Y sin a, X sin a + Y cos a, the axes are rotated in 
the positive direction through an Z a . 



64 ELEMENTARY ANALYTICAL GEOMETRY 

64. By § 35, in an equation of the form x cos a + y 
sin a = p, the length of the J_ from the origin on the 
st. line is the absolute term p. 

If, without changing the direction of the axes, the 
origin be transferred to (x v y^) the equation becomes 

(x + x{) cos a + (y + 2/1) sin a = p, 

i.e., x cos a + y sin « = p — x 1 cos a — y x sin a. 

The new equation is of the same form as the old 
one except that the absolute term is now p — x 1 cos a 

— y x sin a. 

This absolute term is then the length of the _l_ 
from the new origin to the st. line; or, reverting to 
the original origin, the length of the JL from (x v y x ) 
to the line x cos a + y sin a = p is p — x x cos a — y l 
sin a. 

This is the same as the result that would be 
obtained by using the formula of § 45. 

65.— Exercises 

<f! "What does the equation 2a: 2 - llxy -f 12y 2 + 7x - 
13y -f- 3 = become when the origin is changed to the 
point (1, 1) the directions of the axes being unchanged? 

^J2. Transform the equation x 2 -f- xy - 7x - iy + 12 = 
to || axes through (4, - 1). 

3. Find the point that must be taken as origin, the 
directions of the axes being unchanged, in order that the 
terms of the first degree in x and y may vanish from the 
equation x 2 + y 2 + 5x - 9y + 17 = 0. Find also what the 
equation becomes. 



EXERCISES 65 

4. Show that the terms of the first degree in x and y 

will vanish from the expression ax 2 + 2hxy + by- + 2gx + 

/hf ~ bg hg - af\ 
2fy + c, if the origin be changed to (-^- — , -7 — J, 

the directions of the axes being unchanged. 

^6. Transform the equation Ax- + By + C = by rotating 
the axes through an L of 30°. 

6. Find what the equation x 2 - y 1 = a 2 becomes when 
the axes are turned through an L of 45°, the origin 
remaining the same. 

/^. Show that the equation x 2 + y- = a 2 is not changed 
when the axes are turned through any L a, the origin 
remaining the same. 

^8. Find what the equation 33a; 2 - 34 V^xy - y 2 = 
becomes when the axes are turned through an L of 60°, 
the origin remaining the same. 

9. Find the smallest positive L through which the axes 
must be turned in order that the coefficient of xy in the 
equation 59x 2 + 24 xy + 66y' 2 = 250 may vanish ; and also 
find what the equation becomes. 

10. Show that the term involving xy in the expression 
ax 2 -\- 2 hxy -\- by 2 will vanish, if the axes are turned 
through the A 

2 a - b 



66 ELEMENTARY ANALYTICAL GEOMETRY 

6G.— Review Exercises 

1. Find the distances between the following pairs of 
points : — 

(a) (-2, 7), (6, -2); 
(6) (2a + 6, a - 26), (a - b, 3a + b) ; 
(c) (a cos a, a sin a), (-b cos a, - b si/i «). 
Verify the result in (6), on squared paper, when a = 1, 
b = - 2. 

^2; A ( — 5, -1), B (4, 6) are two given points, P is 
taken in AB and Q in AB produced such that AP : PB = 
AQ : QB = 5:3. Find the coordinates of P and Q. 

V J$". Find the area of the A of which the vertices are 
(3a, 2b), (2a, 36) and (a, 6). 

"">£ Find the area of the A contained by the lines 
2x + Uy -f 43 = 0, 9a: + 8y - 14 = and 7x - ty + 
2G = 0. 

jf. Find the _L distance from ( - 2, 3) to the line 

3 2 

Should the result be considered positive or negative and 
why 1 ? 

6. Find the condition that the three points (x v y 1 ), 

( x 2' y-i)' ( ,r 3> v-i) ma y ^ e * n a s ^- ^ me - 

Jfc Find the locus of a point such that the square of its 
distance from ( — 3, - 7) exceeds the square of its distance 
from (5, 0) by 43. 

8. Prove that the equation Ax + By + C = represents 
a st. line. 

9. Find the equation of the st. line which is equidistant 
from the || lines ax -f- by = c, ax -f by = d. 



REVIEW EXERCISES 67 

10. Find the equation of a st. line which makes an Z a 
with Oy and cuts off an intercept b from Ox. 

•^Vt Show that the st. lines Ax -f By + C = 0, A,x + B^ 
■f C, = are ||, if AB X = AiB. 

1*2. Explain the meaning of the constants in the equations 
x - h y - k 
cos sin 6 

^\&. Show that the line y = x tan a passes through the 
point (a cos a, a sin a), and find the equation of the _L to 
the line at that point. 

^]A. Find the Z between the st. line joining ( - 4, 5), 
(5, 1) and the st. line joining (3, 7), ( - 6, - 3). 

^lX Find the values of in and b such that the line 
y = rax -j- b will pass through (3, - 2) and ( — 1, - 5). 

\/y!}. Find the equation of the st. line which passes 
through (2, - 2), and makes an L of 150° with Ox. 

17. Find the length of the st. line drawn from (h, k), in 
the direction inclined at L a to Ox, and terminated in 
the line y = rax -f- b. 

^18. Find the equation of the st. line through (h, k), and 
x v 

it.- + f-i. 

^>9. Show that the st. lines Aa; + By + C = 0, A^ + 
B x y + C x = are _L to eacli other, if AA : -f BBj = 0. 

y^Q. Write the equation of the st. line which is ± ax - 
by — c, and cuts off an intercept = d from Oy. 

J%1. Find which of the following points are on the origin 

side of * -.1 = 1:— (5,3),(-2, -8), (2, -2),(-6, -14), 
3 5 

(-7, -1"). Illustrate by a diagram on squared paper. 



68 ELEMENTARY ANALYTICAL GEOMETRY 

22. Show that the points (2, 6), (1, 11), (-4, 7), (- 3, 3) 
are in the four different angular spaces made by the lines 



+ J-lMld-B-£--l. 



3 T 5 = l and 5 - 9 
Illustrate by a diagram on squared paper. 

23. Find the values of a for which the lines 2x - ay 4 1 
= 0, rt.r - 6?/ - 1 = 0, 18a; — ay — 7 = are concurrent; 

and find also the coordinates of the respective points of 
intersection. 

24. Show that the condition that the lines ax + by = 1, 
cx-\-dy = 1, hx -\- ky = 1 are concurrent is the same as 
the condition that the the points (a, b), (c, d), (A, k) are 
collinear. 

25. Find the L contained by the lines 4x - 7y -f- a = 0, 
3a; 4- ii y 4-fc = 0. 

26. Find the equation of the st. line passing through the 
intersection of Ax - ly -\- a = and 3x + l\y ■+ 6 = 
and making an L of 45 3 with the axis of x. 



27. Find the equation of the st. line passing through the 

y 

b 



intersection oi - -\- £ = 1, y = mx -{- c and also through 



(d, 0). 

28. Show that the equation of the st. line joining the 
intersection of x cos a -\- y sin a = p, x cos /? + y sin {3 = p 

to the origin is y = x tan . 

29. Find the length of the _]_ from (a, b) to - -f | = 1. 

30. Find the equation of the st. line through ( - 5, 1) 
and || to 3.x + Uy = 17. 

31. Find the equation of the st. line through (8, —2) 
and X to 7x = y -f 4. 



REVIEW EXERCISES 69 

32. Find the coordinates of the four points each of which 
is equally distant from the three lines -j- - -_■ = 1, 

12 **" 5 " l ' 24 7 = • 

'"'-33: Find the coordinates of the foot of the _L from 
(3, 5) to the st. line joining (-1, -2) and (8, 1). 

34. Find the separate equations of the st. lines represented 
by 3cc 2 + Uxy -2y* = 0. 

35. Find the product of the J_s drawn from (3, -2) to 
the st. lines represented by ox 2 -f- Vlxy -f 2y 2 = 0. 

36. Show that the L between the lines y — ynx -J- a, 

. - . . . m - n 

y = nx -\- o is tan l . 

1 + mn 

37. Find the tangent of the L between the st. lines 
represented by 5a: 2 - &xy -y 1 = 0. 

t38T Find the equations of the st. lines which pass through 

(3, 6), and are inclined at an L of 45° to — + ^- = 1. 

-337 Show that the st. line joining the point (1, 1) to the 

intersection of f- r- = 1 with - — (- — = 1 passes through 

a b b a 



the 



origin. 



40. Show that the equation of the st. line passing through 
the intersection of x cos a -f- y sin a = >p,x cos /3 -{- y sin ft = q, 
and || to x -\- y = k is 
(x -\- y) sin (a - /S) -f p (si?i /3 - cos /5) -f q {cos a - sin a) = 0. 

^■tfi. Find the equation of the st. line passing through the 
intersection of 5x — ly = 16, 2x - 3y = 7 and J_ to 
6a; -4y = 19. 



70 ELEMENTARY ANALYTICAL GEOMETRY 

42. Find the equations of the st. lines drawn through 
the vertices and || to the opposite sides of the A of which 
the equations of the sides are 3x + Hy = 23, 4x — 9y = 11, 
7x -2y = -31. 

43. Show that the lines 5.x- + y = 4, 2x + y = 2, 3a; + 3?/ 
= 4 are concurrent ; and find the coordinates of their 
common point. 

44. Find the equation of the st. line passing through 
the intersection of ax + by -\- c = 0, fx -4- gy -J- h = 0, and 
(a) also through the origin ; (b) _L to x -f- y — k. 

s4f>. Find the equations of the st. lines which bisect the 
Zs between the lines I2x - 5y = 17, 8a; + 15y = 13. 

-46. Find the equation of the st. line which passes through 
the point of intersection of the lines 5x -|- y = 4, 4x - 9y 
= 11, and is A. to the former. 

47. Show that the points (4, 2), (6, 2), (5, 2+^3) are 
the vertices of an equilateral A. 

48. Find the locus of a point which moves so that the 
sum of its distances from the axes is 10. Trace the locus 
on squared paper. 

49. Find the locus of a point which moves so that the 
difference of its distances from the axes is 10. Trace the 
locus on squared paper. 

50. Find the equations of the st. lines each of which 
passes through ( - 5, - 3) and is such that the part of 
it between the axes is divided at the given point in the 
ratio 7 : 3. 

M. Find the equation of the st. line which passes through 
(3, - 2), and is _L to 4* + y -f 12 = 0. 

52. Find the equation of the right bisector of the st. 
line joining (a, b) and (h, k). 



REVIEW EXERCISES 71 

J&. Two st. lines are drawn through (0, - 3) such tliat 
the J_s on them from ( - 6, - 6) are each of length 3. 
Find the equation of the st. line joining the feet of the ±s. 

54. Find the Z of inclination of the lines ax-\-by = c, 
(a -\- b) x - (a - b) y = d. 

^J>5. A st. line is drawn through (2, - 4) and ± to 7x 

— 3y — 11. Find the equations of the bisectors uf tne Zs 
between the _L and the given st. line. 

56. Find the equation of the st. lines which bisect the 
Zs between the lines represented by x 2 -f- 2xy sec 6 + y 2 — 0. 

57. Find the value of h for which the equation 3x 2 -4- hxy 

— 10y 2 -f- x + 29y -10 = will represent two st. lines. 

58. Show that, if the axes are rotated through an Z of 
45°, the terra containing xy vanishes from the equation 
x 2 -j- 2xy sec d + y 2 = 0; and the separate equations of the 

a 

two st. lines become x = ± y tan — 

&9. Three vertices of a ||gm are (3, 4), (-3, 1), (5, -2). 
Find the coordinates of the fourth vertex. 

60. Prove that the two st. lines which join the middle 
points of the opposite sides of any quadrilateral mutually 
bisect each other. 

61. What must be the value of m, if the line y = mx 
-5 passes through the intersection of 7x - lly = 14 and 
5*4- 2y = -11. 

62. Find the area of the A contained by the lines x f y 
= 12, 2x -y = 12, x -2y = -12. 



CHAPTER IV 

The Circle 

67. A circle is the locus of the points that lie at a 
fixed distance from a fixed point. 

The fixed point is the centre and the fixed distance 
is the radius of the circle. 

68. To find the equation of a circle having- its 
centre at the origin. 




Let P (x, y) be any point on the circle of which 
the centre is O. Let the radius = a. 

Draw PM ± Ox. Join PO. 

*.' OPM is a rt.-zd A, 
.'. OM 2 + PM 2 = OP». 

x 2 + y 2 = a 2 . 

72 



THE CIRCLE 



73 



This being the relation which holds between the 
coordinates of any point on the circle and the given 
radius is the required equation. 

69. To find the equation of a circle, the centre 
being at any fixed point (h, k) and the radius equal 
to a. 



y 




""^p^i) 






/ u-/ ; \ 






( ' ' « 






I M)c 


* ,'l 




O 


h 


j M 


r 



C (h, k) is the centre ; and P (x, y) is any point on 
the circle. 

Draw PM, CN j. Ox, CL j_ PM. Join CP. 

CL = NM = OM — ON = x — h; 

PL = PM - LM = PM - CN = y — k. 

7 CPL is a rt.-^d A, 

CL 2 + PL 2 = CP 2 . 

.-. (x - h) 2 + (y - k) 2 = a 2 . 
This is the required equation. 



74 ELEMENTARY ANALYTICAL GEOMETRY 

7v. If we expand the equation found in § 69, we 
obtain : — 

a? + 2/2 _ 2Jwc - 27cy + h? + k 2 - a 2 = 0. 

Comparing this result with the general equation of 
the second degree : — 

ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0, 

we see that the conditions that the latter should 
represent a circle are that the coefficients of x 2 and y 2 
should be equal and that the coefficient of xy should 
be zero. 

Thus the equation 

ax 2 + ay 2 + 2gx + 2fy ■+• c = 

may be changed to 

V V _i_ /„ j_ /V _ (f +/ 2 -ac 



(• + tr + G' + # 



a- 



from which, by comparison with the formula of § 69, 
we see that it represents a circle having its centre at 

the point (——, — —) and its radius = — 

r V a a' a 

71. The general equation of the circle to rectangular 
axes is commonly written : — 

x 2 + y 2 + 2gx + 2fy + c = 0. 

When the circle passes through the origin and its 
centre is on the axis of x, the equation of § 69 
becomes 

(x - a) 2 + y 2 = a 2 , 

or, x 2 + y 4 = 2ax. 



EXERCISES 75 

72.— Exercises 

-1. Write the equation of the circle with centre (0, 0) 
and radius = i/5. 

w 2. Write the equation of the circle with centre (6, 2) 
and radius = 3. 

v 3. Write the equation of the circle with centre ( - 5, - 1) and 
radius = |/26. Show that this circle passes through the origin. 
4. Write the equation of the circle with centre ( - a, - b) 
and radius = c. Find the condition that this circle passes 
through the origin. 

/&. Find the coordinates of the centre and the radii of 
the following circles : — 

(a), x 2 + y 2 - Qx - 2y = 15 ; -(b), ix- -f 4y 2 4- 7x + 5y 
= 16 ;--(c), x 2 + y 2 = lix-;(d), x 2 + y 2 + 2 by = c 2 . 

6. Draw, on squared paper, the circles of which the 
equations are : — 

(a), x* + y2 = 9 . ( b)j X 2 + y 2 = 8x ; (c), as* + y* + 6y = 7. 

7. Find the centre and radius of the circle which passes 
through the origin and cuts off intercepts = a and b from 
Ox and Oy respectively. 

Solution. — Since the circle passes through the origin its equation 
must be satisfied by x = 0, y = 0, and .". the absolute term must be 
zero. Thus the equation may be written 

x 2 + y 1 + 2 gx + 2/y = 0. 
Substituting in this equation the coordinates of the points (a, 0), 
(0, b) the two equations 

a 2 + 2 ga = 
6 2 + 2 /& = 

are obtained from which a = - - , f = - — . 
" 2 2 

.*. the equation of the circle is 

x 2 + y 2 — ax — by = 0, 

.*(.-i)' t (,-iy-£±JK_ 

.". the centre is ( " — ) and the radius = • 

V.2 2/ 2 



76 ELEMENTARY ANALYTICAL GEOMETRY 

' 8. Find the equation of the circle which passes through 
the origin and also through (4, 3) and ( - 2, G). 

• 9. Find the equation of the circle which has its centra 
on the axis of x and which passes through the points 
(5, 3) and (-3, 1). 

10. Show from the general equation of §71 that three 
conditions are necessary and sufficient to determine a circle. 

11. Find the condition that the circle x 2 + y 1 + %ff x + 
%fy -\- c = may have its centre («) on the axis of x ; 
(b) on the axis of y. 

v 12. Find the equation of the circle which passes through 
(3, 1) and (5, - 3) and has its centre on the line x - y = 4. 

13. Find the equation of the circle having the st. line 
joining (7, - 5) and ( - 3, - 1) as a diameter. 

14. A (a, 0) is a fixed point and P (x, y) is a variable 
point such that PO : PA = p'.q. Show that the locus of P 
is a circle having its centre on Ox, and dividing OA inter- 
nally and externally in the ratio p : q. 

J )£>. Find the equation of the circumcircle of the A whose 
vertices are (3, 4), ( - 2, 3), ( - 5, - 7). 

^>. Find the length of the chord of the circle x 2 + y 2 = 
25 cut off by the line 3x + y = 15. 

^ Vt. Find the length of the chord of the circle x 2 + y 2 - 
6x -f- 14y = 42 cut off by the line x - y = 8. 

18. Show that the locus of the centres of all circles which 
pass through two given points (p, q), (r, s) is the right bi- 
sector of the st. line joining the given points. 

19. Through the given point P (h, k) a st. line is drawn 
cutting the circle x 2 + J/ 2 + 2<?a? -f 2fy + c = at A and EL 
Prove that PA . PB is constant for all directions of the st. 
line. 



EXERCISES 77 

x - h y - k 

Solution : — Take tt = — - — 3- = r as the equation of the st. line. 

cos a sui a ^ 

Then 

x = h + r cos 6, y = k + r sin 6. 

Substituting these values in the equation of the circle, and simplifying 
r 2 + 2 {{h + g) cos d + (k + f) sin d} r 

+ h* + k 2 + 2gh + 2fk + c - 0. 
The value of PA . PB = the product of the two values of r in 
this equation 

= h 2 + ¥ + 2gh + 2/k + c, 

an expression which does not contain and which is .'. independent 
of the direction of the line. 

20. Find the equation of that chord of the circle x- -f- 
y 2 + %9 X + %fy + c = which is bisected at the point (h, k). 

Solution:— (First Method). As in Ex. 19, if we take for the 

as - h y - k 

equation of the chord tt — — : — 7T — r, we get 

^ cos sin i) ' 6 

r 2 + 2 {(A + g) cos 6 + (k + f) sin 6 } r 

+ h* + k 2 + 2gh + 2/k + c = 0. 

If the chord is bisected at (h, k) the two values of r got from this 

equation are equal in value but opposite in sign, and 

.-. {h + g) cos 6 + (k + f) sin 6 = 0. 

Multiplying the terms of this equation by the equal fractions 

, • , the required equation is found to be 

cos sin 6 u H 

(h + g) (x - h) + (k +f)(y- k) = 0. 

(Second Method). Let the equation of the chord be 

y — k = m (x - h). 
The centre of the circle is ( - g, -/), and the equation of the J. 
from the centre to the chord u 

m (y + f) + x + g = 0. 
The _1_ from the centre bisects the chord, and, ,\ passes through 
(h, k). 

:. m{k+f) + h + g = 0. 

_. h + 9 

k +/' 

.". the required equation is 
(k + g) (x - h) + (k + f) (y - k) = 0. 



78 ELEMENTARY ANALYTICAL GEOMETRY 

v 3-1. Find the equation of the chord of the circle x 2 + y 2 - 
6x - 8y — 24 which passes through (5, - 1) and is bisected 
at that point. 

22. From the point P ( - 3, - 7) a st. line is drawn to 
cut the circle x 2 + y 2 - ix - \0y = 17 at A and B. 
Find the area of the rectangle PA . PB. 

23. Find the equation of the common chord of the 
circles 

x 2 + y 2 - Sx - 6y = 39, 

x 2 + y 2 + 6x + % = 56. 

24. Find the condition that the common chord of the 
circles 

x 2 + y i + 2gx + 2/y + c =0, 

x 2 + y 2 + 2g'x + 2/'y + c = 
passes through the origin. 

25. Find the equation of the circle which passes through 
the origin and also through (A, k) and (k, h). 



TANGENTS 



79 



Tangents 
73. Let APQ be a secant cutting a curve at P and Q. 




If the secant rotate about the point P until the 
second point Q approaches indefinitely near to P, the 
limiting position PR of the chord is called a tangent 
to the curve at the point P. 

The point P is called the point of contact of the 
tangent PR. 



80 ELEMENTARY ANALYTICAL GEOMETRY 

74. To find the equation of the tangent to the 
circle Jf 2 + y 2 = a 2 at the point P (x v y Y ) on the circle. 




Fig. 35. 

Let Q (x 2 , y 2 ) be another point on the circle. 
Then the equation of PQ is 

x ~ x i _ V ~Vi 

X l ~ X 2 V\ — Vl 

V P and Q are both on the circle, 



a) 



x \ + U\ — a<2 > 
and, x 2 2 + y 2 = a 2 . 

:. , subtracting, x x 2 - x 2 2 + y* — y 2 2 = 0. 
.*., (x, - x 2 ) {x x + x 2 ) + (y 1 - y 2 ) ( Vl + y 2 ) = (2) 
Multiplying the terms of (2) by the equal fractions 
in (1) 

(x - x x ) (x x + x 2 ) + (y - yj ( Vl + y 2 ) = 0. (3) 

If, now, PQ rotates about p until Q coincides with 
P, x 2 = x 1 and y 2 = y v 

Thus equation (3) becomes 

2 (x - x,) x,+ 2 (y - y x ) y l = 0, 
or, 



TANGENTS 81 

But x x 2 + y x 2 = a 2 . 

■'• x*i + yy x = a 2 . 
This is the required equation. 

75. Alternative Method of finding the equation of 
the tangent at the point P (x x , y x ) on the circle x 2 + 
y 2 = a 2 . 

Using the figure of § 74 let the equation of PQ be 



x-xy y -y x 



(1) 



cos 6 sin 6 
and .'. x = x x + r cos 6, y = y x + r sin 8. 

Substituting these values of x, y in the equation of 
the circle, and expanding 

r 2 + 2 (scj cos 6 + y 1 sin 6) r + x x 2 + 2/j 2 = a 2 (2) 

Since P is on the circle, x x 2 + y 2 = a 2 . 

.'. one value of r is zero, and equation (2) becomes 

r + 2 (x x cos 6 + y x sin 6) = 0. (3) 

If, now, PQ rotates about P until Q coincides with 
P, the other value of r also becomes zero, and, 

.'. x x cos 6 + y x sin 6 = 0. (4) 

Multiplying the terms in (4) by the equal quantities 
in (1), 

x i ( x - Xi) + Vi (y - Vi) = °- 

.-. xx x + yy x = x 2 + y? = a 2 . 
.'. the required equation is 
xx x + yy x = a 2 . 



82 ELEMENTARY ANALYTICAL GEOMETRY 

76. The equation of OP (Fig. 35) is - = V -, and by 

the condition of perpendicularity, the line represented 
by this equation is ± to the line represented by 
xx i + Wx = « 2 - 

.*. the radius of a circle drawn to the point of 
contact of a tangent is J_ to the tangent. 

77. In any curve, the st. line drawn through the 
point of contact of a tangent and _L to the tangent 
is called a normal to the curve at that point. 

78. To find the equation of the tangent to the 
circle x 2 + y 2 + 2gx + 2fy + c = at a point 
p (*i> </i) on the circle. 

Let the equation of a chord PQ be 



V ~ Vi 



(1) 



cos 6 sin 6 

and /. x = x l + r cos 6, y = y 1 + t sin 6. 

Substituting these values of x, y in the equation of 
the circle and simplifying, 

r 2 + 2 { (x l + g) cos 6 + (y 1 + f) sin 6 } r 
+ x 2 + y 2 + 2gx x + 2f Vl + c = 0. 

Since P (x lt y^ is a point on the circle, this equation 
reduces to 

r + 2 { (x x + g) cos 6 + (y 1 + f) sin 6 } = 0. 

If now the secant PQ rotates about P until Q 
coincides with P, the second value of r becomes zero, 
and .-. 

(x, + g) cos + (y> +/) sin = 0. (2) 



TANGENTS 83 

Multiplying the terms in (2) by the equal quantities 
in (1), 

- X x ) (x x + g) + (y- y x ) (y x + f) = 0. 

.'• x ( x i + 9) + V (y L + /) -x* -y x 2 -gx x -fy x = 0. 
But, x* + y x 2 + 2gx x + 2fy x + c = 0. 

.*. , adding, 

xfa + g) + y (y x + /) + gx x + fy x + c = 0. 

» _xxi + yyi + g (x + x t ) + f (y + yj + c = o. 

79. By comparing the equation of the tangent 

xx i + Mi + 9 ( x + x i) + f (V + Vi) + c = o 
with that of the circle 

x 2 + i/ 2 + 2 gx + 2 /y + c = 0, 

the following rule is obtained for writing the equation 
of the tangent at a point (x x , y x ) on the circle : — 

In the equation of the circle change 
x 2 into xx x , 2/ 2 into yy x , 
2x " x + x lt 2y " y + y v 



84 



ELEMENTARY ANALYTICAL GEOMETRY 



80. To find the equation of the tangent to the 
circle x~ + y 2 = a 2 in terms of m, the slope of the 
tangent. 




To find the abscissae of the points where the line 
y = mx + k cuts the circle, eliminate y by substitution, 

and »-t-^*" a 

X 2 + ( mx + ] c y = a \ 

i.e., (1 + m 2 ) x 2 + 2 mJex + lc 2 - a 2 = 0. 

If the line is a tangent, the values of x from this 
equation are equal to each other, and 

;.hn 2 k 2 =/<l + m 2 ) (k 2 - a 2 )r' 

k 2 = a 2 (1 + m 2 ), 



and k = ± ai/1 + on 2 . 
Thus the equation of the tangent is 



y = mx ± a VI + m 2 - 

The double sign corresponds to the two tangents 
that have the same slope, as indicated in the diagram- 



EXERCISES 85 

81.— Exercises 

1. Find the equation, of the tangent to the circle 

(a) x 2 + y 2 --= 34, at the point (3, 5) ; 

(b) x 2 + y 2 - \Qx + \2y = 39 at (-1, 2); 

(c) x 2 + y 2 + 18x - 14y = 39 at (3, 12); 

(d) x 2 + y 2 + 2gx + 2/y =0 at the origin. 

--2r Find the equations of the st. lines touching the circle 
x 2 + y 2 = 35 and making an L of 45° with the axis of a;, 

J^3T Find the equations of the st. lines which touch x 2 + 



/£ Prove that x + 2y = 10 is a tangent to the circle 
x 2 + y 2 = 20 ; and find the point of contact 

\8. Prove that x - 2y = 4 is a tangent to the circle 
x 2 + y 2 - 8x - lOy + 2\ =0; and find the point of 
contact 

G. Find the condition that Ax + By + C = may touch 
(a) x 2 + y 2 = r 1 ; (b) x 2 + y 2 + Igx + 2/y + c = 0. 

7. Find the condition that — + f- — 1 may touch 
9,90 a b 

x L + y J = r-. 

8. Find the condition that the axis of x may touch 
* 2 + y 2 + 2(/x + 2/y + c = 0. 

9. Find the equations of the circles passing through 
(5, 2) and touching the axes of x and y. 

^Kh Find the equations of the tangents to the circle 
x 2 + y i _ 2x + 2y = 10 which make an L = 30° with 
the axis of a:. 



86 ELEMENTARY ANALYTICAL GEOMETRY 

11. Find the length of the part of the line 2x - by + 
10 = intercepted by the circle x 2 + y 2 + 8x - 6y = 24. 

12. Show that the tangent to the circle (x - a) 2 + 
(y - b)' 2 = r 2 at the point (x v y x ) on this circle is (x - a) 
(*! - a) + (y - b) (y x - b) = r 2 . 

Solution. — Transform the origin to the point (a, b) without changing 
the direction of the axes. The transforming relations are x = X -|- a, 
y = Y + b, X,. = X l + a, Vl = Y 1 + 6. 

The equation of the circle becomes 

X 2 -f Y 2 = r\ 
and, by § 74, the tangent at (X 1; Y x ) is 

XX, + YY, = r\ 

Transforming back to the original origin, the equation of the tangent 
becomes 

(x - a) (x, - a) + (y - b) (y, - b) = r\ 

13 Show that the point (g + r cos a, f + r sin a) is on 
the circle (x - g) 2 + (y - y*) 2 = r 2 ; and find the equation 
of the tangent at that point. 

14. Show that the circles 

x 2 + y 2 + 6x + IGy + 24 = 0, 

a; 2 + y 2 - 10* + iy + 20 = 

touch each other externally. Find the coordinates of the 
point of contact ; and the equation of the common tangent 
at that point. 

15. Show that, if the circles 

(x - hf + (y - k) 2 = r\ 
(x - mf + (y - n) 2 = s 2 
touch each other, 

(h - m) 2 + (k - n) 2 = (r ± s) 2 . 



EXERCISES 87 

_^16: Find the equation of the common chord ; and the 
coordinates of the points of intersection of the circles 

x 2 + y 2 - 2x - 6y = 14, 
x 2 + y t _ 5 X + 3 y = 5. 

"17. Find the equation of the circle who«e centre is at 
the origin and which touches the line A.v + By + C = 0. 

JL^f Find the equation of the circle whose centre is at 
(7, 2) and which touches 3.k - 5y = 4. 

19. Find the centres of similitude and the equations of 
the transverse and direct common tangents of the circles 

x 2 + y 2 - 6x + 2y 4- 6 = 0, 
x 2 + if + Sx - lOy + 32 = 0. 

20. Find the equations of the common tangents of the 
circles 

x 2 + y 2 - 4x - Sy - 5 = 0, 

x 2 + y 2 - 10.x - 6y - 2 = 0. 

Find also the coordinates of the points of contact of the 
tangents. 

21. Find the equations of the tangents to the circle 
a? 2 + y 2 + 2yx + 2/y + c = which are || to x + 3y = 9. 

y 22. Find the equations of the two tangents to the circle 
x 2 + y 2 = 25 which make an L of 30" with the axis of x. 






88 



ELEMENTARY ANALYTICAL GEOMETRY 



Poles and Polars 

82. To find the equation of the chord of contact 
of tangents drawn from an outside point to the 
circle x' 1 -f- y~ = a 2 . 




Let P (x v y x ) be the given point; PA and PB the 
tangents. 

It is required to find the equation of AB. 

Take (x, y'), (x", y") to represent the coordinates 
of A, B respectively. 

The equation of AP is, by § 74, ' 

xx + yy' » a 2 ; 

and since the coordinates of P must satisfy this 
equation, 

*& + VxV = a 2 - (1) 



Similarly, x^x" + y x y" = a 2 



(2) 



POLES AND POLARS 89 

From these results it is seen that 

® x i + 2/2/i = « 2 
is the equation of AB; for: — 

since it is of the first degree it represents a st. line; 
by (1), A is a point on the line ; 
by (2), B is a point on the line; 
.'. the required equation is 

xx x + yy x = a 2 . 



83. The equation of the chord of contact of tangents 
drawn from an outside point to a circle is of the same 
form as the equation of the tangent at a point on the 
circle. 

This is in agreement with the fact that, if the point 
P approach the circle and ultimately fall on it, the 
chord of contact becomes the tangent at P, or the 
tangent at P is the final position of the chord of 
contact when p approaches the circle. 



90 



ELEMENTARY ANALYTICAL GEOMETRY 



84. To find the equation of the polar of P (x v yj 



with respect to the circle x- + y~ 



a\ 




Through P draw any st. line cutting the circle at 
A, B. Draw tangents AQ, BQ intersecting at Q (X, Y). 



It is required to find the locus of Q. 



r ^ f 



By § 82, the equation of a£ is 
xx + yy = a 2 ; 
and as the coordinates of P must satisfy this equation 
Xx 1 + Yy x = a 2 . 

:. , as X, Y are the coordinates of any point on the 
polar of P, the required equation is 

xx x + yy : = a 2 . 

85. The equation of OP is xy x — yx x = 0, and, by 
the condition for perpendicularity, this line is _|_ to 
that represented by xx x + yy x = a 2 . 

.'. the polar of P is a st. iine which cuts OP at 
rt. Ls 



POLES AND POLARS 91 

86. If the polar xx x + yy x = « 2 cuts OP at M, the 

2 

length of OM = — = _ . 

.'. OM . OP = a 2 . 

87. The equation of the polar of the point P (x x , y x ) 
without the circle x 2 + y~ = a 2 is the same as that of 
the chord of contact of tangents drawn from P to the 
circle. This shows that, when the point is without 
the circle, its polar is the chord of contact produced, 
or, that tangents drawn from P touch the circle at 
the points where it is cut by the polar of P. 

88. The equation of the polar of any point P (x v y x ) 
is of the same form as the equation of the tangent 
at a point on the circle. 

This is in agreement with the fact that, if the 
point P approaches and ultimately coincides with the 
circle, OP becomes equal to a, and .'., by § 86, OM 
becomes equal to a, and the polar becomes the tangent 
at the point P 



92 ELEMENTARY ANALYTICAL GEOMETRY 



80. If the polar of P passes through Q, the polar 
of Q passes through P. 

Let (x lt 2/i)> ( ,x 2» 2/2) ^e ^he coordinates of P, Q 
respectively. 

The polar of P with respect to x 2 + y 2 — a 2 is 

xx x + yy x = a 2 . 
Since this line passes through Q, 









aw + y&\ = a ~- 



il^i 



This proves that (x v y x ) is on the line 

^2 + yy-2 = « 2 ; 

and .'. P is on the polar of Q. 

Cor. If the point Q moves along the polar of P, 
the polar of Q changes its position, but always passes 
through P. 

.*. , if the pole moves along a st. line, its polar 
turns about the pole of that line. 

90. To find the pole of the st. line ax + By + c = 
with respect to the circle x 2 + y 2 = a 2 . 

Let (x v y x ) be the coordinates of the pole- 

The equation of the polar of (x v y x ) ls 

xx x + yy x — a- — 0. 
This equation must be the same as 

Ax + By + C = 0. 



x A 
A 

X X = 



Vi — Q? 




B ~ ~C~ 




c ' yx 


a 2 B 



POLES AND POLARS 93 

91. To find the polar of P (x v y Y ) with respect 
to the circle x- + y 2 + 2gx + 2fy + c = 0. 

The equation of the circle may be written 

(* + i/) 2 + (y + /) 2 = i/ 2 + / 2 - & 

Transforming the origin to the point (— g, — /'), the 
transforming relations are x = X — g, y — Y — /, 
x 1 = X x — #, ^/j = Yj — /, and the equation of the 
circle becomes 

X 2 + Y 2 = g- + f - c 

The equation of the polar of P (X„ Y : ) with respect 
to this circle is, by § 84, 

XX X + YY X = g- + p - c. 

Transforming back to the original origin, the equation 
becomes 

(x + g) (x, +g)+(y+ f) ( yi + f) = g* + f* - c. 

'• xx x + yy t + g (x + Xl ) + f (y + y x ) + c = 0. 

Note. — As an exercise, the student should obtain the above 
result directly from the definition of poles and ])olars, by 
the method used in § SJ/.. 



94 ELEMENTARY ANALYTICAL GEOMETRY 

92.— Exercises 
X. Find the polar of the point 

(a) (3, 5) with respect to x 2 + y 2 = 30; 

J (b) (a, 0) .. n< ii a: 2 + y 2 = r 2 ; 

(c) ( - 2, 4) ,. ii n .x 2 + y 2 - 4.x - &y = 5 ; 
(rf) (-5, -1) ii ii .. x 2 + f - 10.x + 6y = 15; 
(«) (0, 0) (.x - h) 2 + {y - h) 2 = r 2 . 

2. Find the pole of the st. line 

(a) Ix - 7y = 17 with respect to a; 2 + y 2 = 17; 
(6) x - 1y + 12 = i- ., „ x 2 + y 2 = 23; 

>- (c) 4.x- - y = 1 with respect to x 2 + y 2 - Ix— 4y = 4; 
(rf) 4.x + 5y = 5 .. ii „ .x 2 + y 2 - 8.x - lOy 

= - 5. 
Joy(a) Show that .x 2 + y 2 = 25 is the equation of a circle. 
J(b) Show that ( - 3, 4) is on the circle. 

• (c) Write the equation of the tangent to the circle at 
this point. 

(d) Show that the point (9, 13) is on this tangent. 

(e) "Write the equation of the polar of (9, 13). 

(j) Find the equation of the st. line through (9, 13) 
_L to the polar, commenting on the form of the result. 
(g) Find the equation of the other tangent from (9, 13). 
Draw the diagram on squared paper. 

4. Find the pole of — + |- = 1 w i t h respect to x 2 + y 2 = c 2 . 

5. Find the pole of Ix + my = 1 with respect to the 
circle a~ + y 2 + 2gx + 2fy + c = 0. 

G. Prove that the polar of ( — 2, 5) with respect to x 2 + 
y 2 = 18 touches x 2 + y 2 - G.c + 2y = 19; and find the 
coordinates of the point of contact. 



tangents from an outside point 95 

Tangents from an Outside Point 

93. To find the length of the tangent PA from the 
point P {x v y^ to a given circle. 




Fig. 40. 

(1) Let the equation of the circle be 

x 2 + y 2 := a' 2 . 
Join OA, OP. 



V AOP is a rt.-Zd A, 
AP 2 = OP' 2 - AO 2 

= x x + Vi ~ « 2 . 

.'. ap = /xg + v/ - a 2 . 
(2) Let the equation of the circle be 

x 2 + y 2 + 2gx + 2fy + c = 0. 
This equation may be written 

(» + #/ + (y + /) 2 = £ 2 + f - c, 
from which it is seen that the centre is ( — (J, —f) 
and the radius = Vg 2 + f- — c. 

"With the diagram and construction of Fig. 40, 
AP 2 = OP 2 - AO 2 

= (®i + Of + (2/i + /) 2 -Q/ 2 + f ~ o) 
= x* + y 2 +2 gx x + 2/y, + c. 



AP = r/Xf + y x 2 + 2gx t + 2fy! + c. 



06 ELEMENTARY ANALYTICAL GEOMETRY 

94. To find the equation of the tangents from 
(x x> y x ) to the circle x 2 + y 2 = a 2 . 




Let a secant di^wn from P (x v y x ) cut the circle at 
A ; and let Q (x, y) be any point on the secant. 

If PA : QA = k : 1, the coordinates of A are 
(—j- - 1 , -| ^Y and ."., since A is on the circle 

(*» - ^) 2 + (% - y,f = (* - i) 2 « 2 > 

or, (« 2 + y 2 — a 2 ) I* 2 — 2 («« 1 + 2/2/ x — a 2 )k 
+ x \ + 2/i 2 _ ft2 = °- 

If, now, the secant turn about P until it coincides 
with either of the tangents from P, the two values of 
k found from this equation, and which correspond to 
the two points where the secant cuts the circle, are 
equal to each other. 

/. (xx,. + yyj - a 2 ) 2 = (x 2 + y 2 - a 2 ) ( Xl 2 + y x 2 - a 2 > 

This is the required equation. 



EXERCISES — RADICAL AXIS 97 

95.— Exercises 

1. Find the length of the tangent from 
J (a), (7, 3) to a- 2 + y 2 =, 22 j 

J(b), (3, -5) to x 2 + y 2 - 3x + 1y + 35 = 0; 
(c), (,-2, -6) to a- 2 + y 2 = I2x ; 
(J), (0, 0) to x* + 2/2 + 2 gx + 2/y + c = 0; 
(e), (4, 2) to x 2 + y 2 - 6x + 2y - 39 = 0. 
Explain the imaginary result in (e). 

2. The length of the tangent drawn from a point to 
a: 2 + y 2 - 10a; - 4y + 9 = is always 4. Find the locus 
of the point. Plot the diagram on squared paper. 

3. The length of the tangent from P to x 2 + y 2 = 9 is 
twice the distance from P to (6, 0). Find the locus of P. 
\Jrt Find the equations of the tangents from (7, - 1) to 
a 2 + y 2 = 25. 

5. Show, by the method of § 94, that the equation of the 
tangents from (x v y^) to a 2 + y 2 + 2gx + 2fy + c = is 

{xx x + yy x + g (x + x~) + /(</ + ft) + c} 2 
- (a? + y 2 + 2«^ + 2/y + c) (x 2 + y 2 + 2gx x + 2/^ + c). 



Radical Axis 
96. To find the radical axis of the circles 
x 2 + y 2 + 2gx + 2fy + c =0, 
x 2 + y 2 + 2g'x + 2f y + c' = 0. 
Since the tangents to the circles from any point on 
their radical axis are equal to each other, if (x, y) is 
any point on the locus, by § 93, 

x 2 + y 2 + 2gx + 2/y + c = x 2 + y 2 + 2g'x + 2fy + c' 
.". the required equation is 

2 (g - g) x + 2 (f - f ) y + c - c' = 0. 



98 ELEMENTARY ANALYTICAL GEOMETRY 

97. The centres of the circles in the last article are 
(-g, -/) and (-</, -/')• 

.". the st. line joining the centres is 

+ g = y + f 

9-9 f~f 
or, (/ - /) (0 + g) - (g - g') (y + /) = 0. 

By the condition of perpendicularity this line is j_ to 

2 (g - 9') * + 2 (/-/') y + c - c = o. 

.-, the radical axis is J_ to the line of centres. 

98.— Exercises 

1. Find the radical axis of the circles 

x .2 + y i _ i x _ 6y + 9 = 0, 
x i + y i - 16a; - 14y + 104 = 0. 
Draw the diagram on squared paper. 

2. Find the radical axis of the circles 

2x 2 + 2 2 y + 9x - 8y - 3 - 0, 
a;2 + y 2 __ 9. 

3. Show that the radical axes of the circles 

x 2 + y 2 + Igx + 2/?/ + c =0, 
x 2 + y 2 + 2^ + 2/ x j/ + c x = 0, 
a: 2 + j/ 2 + 2# 2 * + 2f 2 y + c 2 = 

taken two and two are concurrent. (The point of concur- 

rence is the radical centre.) 

4. Find the radical centre of the circles 

x 2 + y 2 - Zx + 1y + 35 = 0, 
x 2 + y- - 7x + by - 31 = 0, 
x 2 + y 2 - 6x + 2y - 39 = 0. 



MISCELLANEOUS EXERCISES 99 

5. Show that the circles 

x 2 + y- - 3x + by - 9 = 0, 

x 2 + y 2 + 7x + y - 11 = 0, 

x 2 + y 2 + 2x + 3y - 10 = 0, 
have a common radical axis. Show also that their centres 
are in a st. line which is J_ to the common radical axis. 



Miscellaneous Exercises 
(a) 

.1. Find the equation of the st. line passing through the 
intersection of x - 2y = 5, x + 3y = 10 and || to 3a; + 
4y = 11. 

%. Find the equation of the st. line passing through the 
intersection of 8x + y = 7, l\x + 2y = 28 and _L to the 
latter line. 

<3. Plot the quadrilateral (4, 2), (-5, 6), (-9, -6), 
(7, - 4) ; and find its area. 

V>/piot the lines 2x + 5y = 29, 12ar + y = 29, 5x - 2y 
= 29 ; and find the area contained by them. 

5. Find the equation of the s\ line passing through 
(h, k) and such that the portion of it between the axes is 
bisected at the given point. 

6. Find the equation of the st. line passing through 
(h, k) and (a) || to, (b) _L to the st. line joining (x v y x ), 
(a* V-z)- 

7. Show that, if the lines ax + by -fr- c = 0, bx + cy + a 
— 0, ex + ay + b = are concurrent but not coincident, 
then a + b + c = 0. 

^8. Find the ratio in which the st. line joining ( - 5, 3), 
(6. - 1) is divided by x - ll.y + 3 = 0. 



100 ELEMENTARY ANALYTICAL GEOMETRY 

.9. Find the centre of the inscribed circle of the A formed 
by the lines Ax - 3y = 18, 5a; + 12y = 9, 24a: + 7y = 30. 

10. Find the area of the A contained by y = 3x, y = 5x 
and x + 1y = 77. 

11. Show that the area of the A contained by y = m^x, 
y = m x and Ax + By + C = 

(m l - ra„) C 2 
~~ 2 (A +~m 1 B) (A~+ w 2 B)' 

12. Find the locus of a point such that the square of 
its distance from (fi, 0) is three times the square of its 
distance from (2, 0). 

13. One vertex of a ||gm is at the origin and the two 
adjacent vertices are at (a, b), (c, d). Find the fourth 
vertex. 

14. Show that, if the two circles 

X 2 + y i + % JX + 2/y + c = 0, 
X 2 + y 2 _ 2/x - 2gy + c = 
touch each other, then (y - /) 2 = 2c. 

15. Give the geometrical interpretation of the equation 
cc 2 + y 2 + 2aa; cos a + 2ay sin a + a 2 = 0. 

16. Find the locus of the intersection of the st. lines 
which pass through (6, 0) and (0, 3) respectively and cut 
each other at rt. Zs. 

17. Find the equations of the tangents to the circle 
jc 2 + y 1 = Ikx which are || to 3x — y = 0. 

18. Find the orthocentre of the A whose sides are 8a; — 
5y = 16, 1x - 3y = 10 and x + 2y = 6. 

19. Prove that the radical axis of the circles 

x 2 + y 2 = « 2 , 

x 2 _|_ yi _ 2a (x cos a + y sin a) = 
bisects the st. line joining their centres. 



MISCELLANEOUS EXERCISES 101 

20. Chords of the circle x 2 + y 2 = a 2 pass through the 
fixed point (h, o). Find the locus of their middle points. 

21. Find the equation of the circle which passes through 
the origin and makes intercepts a and b on Ox, Oy 
respectively. 

22. Find the equation of the circle described on the st. 
line joining the origin to (g,f) as diameter. 

23. Find the coordinates of a point such that the st. line 
joining it to (4, -3) is bisected at rt. Z-s by 2x - 3y = 7. 

' 24. Find the locus of the points from which tangents 
drawn to x 2 + y 2 = 13 and x 2 + y 2 - 2x + 6y + 1 = are 
as 5 is to 3. 

25. Find the distances from the point (2, 4) in the 

direction having the direction cosines - — , - — ^ Q ^ ne 
curve whose equation is 

3a; 2 - 6xy + 5y 2 - 32 = 0. 

26. Find the equation of the locus of a point P such 
that PA : PB = k : 1 where A (x v y x ), B (x 2 , y 2 ) are fixed 
points. 

Show that the locus is a circle and find the relation of 
its centre to A and B. 

^fl ^{(t) Find the coordinates of the point C which 
divides the st. line joining A (3, -2), B (19, 10) in the 
ratio AC : C3 = 1:3. 

(b) Prove that D (11, 4) lies on the st. line AB given 
above; and by computing the lengths of AD and BD, find 
the ratio in which D divides AB. 

28. (<t) Find the area of a A the coordinates of whose 
angular points are (a^ yj, (x 2 , y 2 ), (x. A , y 8 ). 



102 ELEMENTARY ANALYTICAL GEOMETRY 

(6) From the result of (a) deduce the equation of a st. 
line in terms of the coordinates of two given points through 
which it passes. 



29. Show that y = mx + a \/l + m' 2 is always a 
tangent to the circle x 2 + y 2 = a 2 . 

30. The equation 3x 2 + 3y 2 - 12a; - 6y + 4 = can 
be reduced to one containing terms in x and y of the 
jecond degree only, by transforming to || axes through a 
properly chosen point. What are the coordinates of the 
point ? 

31. Find the distance of the point of intersection of the 
lines 3.x + 2y + 4 = and 2x + by + 8 = from the 
line 5x - 12y + 6 = 0. 

32. Find the equation of the circle whose centre is (h, k) 
and which passes through (a, b). 

33. Find the locus of the points from which tangents 
drawn to the circle 

x 9 - + 7/2 + 2gx + 2/y + c = 
are at rt. Z_s to each other. 

34. If the tangents at the points (x v y^), (x 2 , y. 2 ) on the 
circle x 2 + y 2 + 2gx + 2/y + c = are at rt. Ls to each 
other, show that 

^2 + y\Vi + 9 ( x i + x 2) +/(vi + y-i) + 9 2 +f 2 = °- 

35. Find the equation of the st. lire joining (ab 2 , 2ab) 
and (ac 2 , 2ac). 

3G. Show that, the equation of the J_ to — . — x + — - — . y 

a b 

= 1 at the point (a cos a, b sin a) is x — — y 

„ TO cos a sxn a 

= a- — b z . 

37. Find the product of the Is from (-7, - 4) to the 
lines 3a;- - 1 2xy + 1 1 y' 2 = 0. 



MISCELLANEOUS EXERCISES 



103 



38. Show that the product of the J_s from (c, d) to the 
iines ax 2 + 2 hxy + by 2 = is 

ac- + 2hcd + b<P 



39. Find the equation of the st. lines which join the 
origin to the points of intersection of 

ax + by = k (1) 

and x 2 + y 2 + 2gx + 2/y + c = 0. (2) 

Solution : — 




Let the line (1) cut the circle (2) at A, B. 

It is required to find the equation representing OA and OB. 

From (1) k 2 = k (ax + by) = (ax + by) 2 . 

¥(x x + j/ 2 ) + 2* (ax + by) (gx + fy) + c (ax + by? = 0. (3) 

Equation (3) has all its terms of the second degree in x and y, 
and .'. , by § 56, it represents two st. lines passing through the 
origin. 

Again, equation (3) is satisfied by the values of x and y which 
satisfy both (1) and (2); 

.'. the lines represented by (3) pass through A and B. 
.♦. equation (3) represents OA and OB. 



104 ELEMENTARY ANALYTICAL GEOMETRY 

40. Find the equation of the st. lines joining the origin 
to the points of intersection of 2x — Zy = 1 and x 1 + y- = 5. 

41. Find the equation of the st. lines joining the origin 
to the points of intersection of x + 2y — 2a and 5 (.r 2 + y 2 ) 
+ 5ax + l n ay = 18a 2 , and show that they are _L to each 
other. 

42. Find the L between the .st. lines which join the 
origin to the points of intersection of — _ J- = \ an j x i _^_ 
y* - 2x + %y + 1 = 0. 

43. Find the equations of the st. lines passing through 
the intersection of 3x + 2y = 7 and x + 5y =11 and such 
that the J. on each of them from ( 1, 7) is equal to 5. 

44. A, B are points on Ox, Ox' respectively and on 
OA, OB squares OACD, OBEF are described. EF produced 
cuts AC at G. Prove that OG, BC, ED are concurrent. 

45. If 1 is constant, show that the variable line 

a b 
x y 
b -r = 1 passes through a fixed point. - 

46. A st. line moves so that the sum of the J_s to it 
from (a, b), (c, d) is equal to the JL to it from (g, h). 
Show that the st. line passes through a fixed point and 
find the coordinates of the point. 

47. Prove that the difference of the squares of the tangents 
from (x v y x ) to the circles 

* 2 + y- + lg x x + 2 f\V + c i = °» 

x- + y 1 + 2(j 2 x + 2f 2 y + c 2 = 

is equal to twice the rectangle contained by the distance 

between the centres of the circles and the length of the J_ 
from (x v y x ) to their radical axis. 



MISCELLANEOUS EXERCISES 105 

48. Three circles touch each other at a common point. 
Prove that the polars of a fixed point (x v y x ) with respect 
to these circles are concurrent. 

49. Find the equations of the st. lines which divide the 
Zs between the lines Ax - 3y + 7 = 0, 5r + 12y - 19 = 
into parts whose sines are as 5 to 7. 

50. Show that the equation of the st. line joining 
la cos (a + /?), b sin (a + ft\ and la cos (a - ft), b sin (a - ft) I 

is — cos a + ~- sin a = cos ft. 
a b 

51. Show that the bisectors of the interior Zs of a A 
are concurrent. 

Note. — Take the origin within the A, and let the equations 
of the sides be 

x cos a 1 + y sin a 1 = j> v 
X cos a., + y sin a., = p 2) 
x cos a 3 + y sin a % = j> x 

52. If the chord of the circle x 1 + y 1 = a 2 whose equa- 
tion is px + qy = 1 subtends an Z of 45° at the origin, 
then a 2 {f- + q 2 ) = 4 - 2 V2. 

53. A st. line moves so that the sum, or the difference, 
of the intercepts cut off from the axes varies as the area 
of the A contained by the st. line and the axes. Prove 
that the st. line passes, in either case, through a fixed point. 

54. Show that the area of the A contained by the lines 
ax 2 + 2hxy + by 2 = and A.*; + By + C = is 

C 2 / h 2 - ab 



A 2 6 - 2 ABh + B 2 a 

55. OACB is a ||gm, P is a point in OA, Q is a point in 
OB; PS drawn || OB meets BC at S; QR drawn || OA 
meets AC at R. Show that PR, QS, OC are concurrent. 



106 ELEMENTARY ANALYTICAL GEOMETRY 

56. P is a point such that the sum of the _|_s from P 
on Ox and on x - by = is constant. Prove that the 
locus of P is the base of an isosceles A of which O is the 
vertex and y = 0, x — by = are the sides. 

57. Given the base of a A in magnitude and position 
and the magnitude of its vertical Z ; prove that the locus 
of its vertex is a circle. 

5S. Prove that, if (x v y x ), (x 2 , y 2 ) are the extremities of 
the diameter of a circle, the equation of the circle may be 
written 

(x - Xj) (a; - x 2 ) + (y - y x ) (y - y 2 ) = 0. 

59. If (h, k) is a point in the first quadrant, show that 

the equation of the st. line which passes through (h, k) 

and makes with the axes in that quadrant the A of 

.xy 
minimum area is — + £- _ 2 
h k 

GO. Show that, if the chord of contact of tangents drawn 
from the point (A, k) to the circle x" + y 2 = r 2 subtends a 
rt. L at the centre, then h 2 + k' 2 = 2r 2 . 

CI. P, Q are two points and O is the centre of a circle. 
PM is _L to the polar of Q with respect to the circle, 
and QN is _L to the polar of P. Show that PM : QN = 
OP :OQ. 

62. Tangents PA, PB are drawn from the point P (h, k) 
to the circle x 2 + y 2 = r 2 . Prove that 

APAB = r (*' + * 8 f -, rS £ 

h 2 + k 2 

63. Prove that the polar of (a, b) with respect to 
x 1 + y 2 = c 2 is a tangent to (a; - h) 2 + (y - k) 2 = r 2 , if 
(ah + bk - c 2 f = (a 2 + b 2 ) r 2 . 



MISCELLANEOUS EXERCISES 107 

64. ABC is a A in which a variable line DE drawn || 
to BC cuts AB at D and AC at E. Show that the 
locus of the intersection of BE, CD is the st. line joining 
A to the middle point of BC. 

65. Show that the equation of the system of circles which 
pass through (h, k) and touch Ax + By + C — may be 
written 

(Ah + Bk + C) {(Ax- + By + C) 2 + (B.r - Ay + /) 2 } 
= (Aas + By + C) {(Ah + Bk + C) 2 + (Bh - Ak + I) 2 }, 
where I is an arbitrary constant. 

66. The circle x 2 + y 2 + Igx + 2/y + c = cuts off 
from Ox, Oy chords of which the lengths are respectively 
a and b. Show that 4g 2 - a 2 = if 2 - b 2 = 4c. 

67. Find the locus of the middle points of the chords of 
the circle x 2 + y 2 = a 2 which pass through the fixed point 
(h, k). 

68. O is the centre of a fixed circle, A is a fixed point, 
Q is any point on the circle. The bisector of Z AOQ 
meets AQ at P. Show that the locus of P is a circle 
having its centre in AO. 

69. Find the equation of the circle with its centre on 
Ox and which cuts x 2 + y 2 = 9 and 5 (x 2 + y 2 ) = 9x 
orthogonally. 

70. Show that the circles x 2 + y 2 + 2x - 1y = 23 and 
7 (x 2 + y 2 ) - 192a- + 144y = 175 cut orthogonally at the 
point (3, 4). 

71. If the chord of the circle x 2 + y 2 = r 2 on the line 
V 7 ^ + 9V = 1 subtends an L of 45° at the origin, theu 

{r 2 (p 2 + q 2 ) - 4} 2 =8. 



108 ELEMENTARY ANALYTICAL GEOMETRY 

72. A rt. L is subtended at the origin by the chord o{ 
the circle (x - A) 2 + (y — k) 2 = r 2 on the line x cos a + 
y sin a = p. Show that 2p 2 — 2p (h cos a + k sin a) + 
h 2 + k 2 = r 2 . 

73. Show that the condition that the circles x 2 + y 2 = r 2 , 
x 2 + y 2 + 2 gx + 2 fy + c = touch each other is 

(r 2 + c) 2 = 4r 2 (g 2 + f 2 ). 

74. Find the z_ between the tangents at a point of 
intersection of the circles x 2 + y 2 - Ax - 8y = 5 and 
x 2 + y 2 - 10;c - 6y - 2. 

75. The equal sides OA, OB of an isosceles rt.- /.d A are 
produced to P, Q such that AP . BQ = OA 2 . Show that 
PQ passes through a fixed point. 

76. Find the locus of a point such that a tangent drawn 
from it to the circle x 2 + y 2 - 8x - lOy = 8 is twice a 
tangent drawn from it to x 2 + y 2 = 25. 

77. Find the equation of the circle which passes through 
the points of intersection of x 2 + y 2 = a 2 , x 2 + y 2 = 
2a (x + y), and touches the line x + y — 2a. 

78. Show that, if the line — = —. — - = r cuts the 

cos sin 

circle x 2 + y 2 = a 2 at D, E and the polar of the point 
P (A, k) with respect to the circle at F, then P, D, F, E 
is a harmonic range. 

What is the general statement of this proposition ? 

79. If axy + bx + cy + d = represents two st. lines, 

show that the lines intersect at the point ( - — , - — )• 

a a 

80. If ax 2 + 2hxy + by 2 + 2gx + 2fy + c = represents 
two st. lines, show that the squares of the coordinates of 

the intersection of the lines are and 9- - 

h 2 - ab h 2 - ab 



MISCELLANEOUS EXERCISES 109 

81. The st. lines y + mx = b, y + mx = c cut the axes 
at A, B and A', B' respectively. Find the area of AA'B'B. 

82. Show that the polar of (k, h) with respect to the 
circle which has its centre at (h k) and touches the line 
Ix + my + 1 = is (l 2 + m-) (A - k) (y - x + h -k) = (lh + mk + 1 )' 2 . 

83. Prove that the locus of the middle point of the 
chord of contact of tangents drawn from points on a given 
st. line to a given circle is a circle passing through the 
centre of the given circle and having its centre on the 
X from the centre of the given circle to the given st. line. 

84. Three concentric circles, A, B, C, have their radii in 
G. P. Shuw that, if the pole with respect to B of a st. 
line is on A, the polar will touch C ; and if the pole is on 
C, the polar will touch A. 

85. Find the equation of the bisectors of the angles 
contained by the lines x 2 + y 2 + kxy = 0. 

86. Find the locus of a point such that the _L from it 
to the line x + y = a is the geometrical mean between 
the coordinates of the point. 

87. Show that the circles x 2 + y 2 + 2yx + 2/y = 0, 
x 2 + y 2 + %fx - 2<jy = cut orthogonally. 

88. Find the equation of the circle which cuts ortho- 
gonally each of the circles : — - 

x 2 + y 2 + x + 2y = 3, x 2 + y 2 + Sx + iy = 7, x 2 + y 2 + 4x + iy = 8. 

89. Points A, B are given in Ox ; C, D in Oy such that 
OA, OB, OC, OD are in H. P. Show that the locus of 
the intersection of AD and BC is x = y. 

90. Show that the lines x + 13y = 0, 3x = by, 5x = 7y, 
and Ix = 8y form a harmonic pencil. 

91. The circle x- + y 2 = r 2 cuts Ox', Ox at C, D 
respectively. EF is a chord such that L EOF = 2a, and 
CE, DF intersect at P. Show that the locus of P is the 
circle X 2 + y 2 _ 2ry tan a = r 2 . 



110 ELEMENTARY ANALYTICAL GEOMETRY 

92. Find the equation of a circle passing through the 
intersections of the circles : — 

(1) x 2 + y 2 - ix - 8y = 28, (2) a? + y* = 9 ; 
and through the centre of (1). 

93. Show that the points (1, 7), ( - 2, 8), (3, 3) and 
(2, G) are concyclic. 

94. From any point A in the line x = y st. lines are 
drawn making /.s of 60° and 120° with Ox and cutting 
y'Oy at B and C respectively. From OC a part OD is cut 
off = OB. Show that CD = the diagonal of a square 
on OA. 

95. D, E are respectively points in two given st. lines 
OX, OY such that OD + OE = c ; and P is a point in DE 
such that DP = m, EP. If OX, OY are taken as axes 
of coordinates, show that the locus of P is (m + 1) 
(mx + y) = cm. 

96. A point P moves such that the distance of P from a 
'fixed point equals the tangent from P to a fixed circle. 
Show that the locus of P is a st. line _L to the st. line 
joining the fixed point to the centre of the circle. 

97. Show that the st. lines represented by bx 2 - 2/ixy + 
ay 2 = are respectively J_ to the st. lines represented by 
ax 2 + 2hxy + by 2 = 0. 

98. A series of circles touch the axis of x at the origin. 
Show that the tangents at the points where the line y = b 
cuts the circles all touch the fixed circle x 2 + y 2 = b 2 . 

99. A circle of given radius moves so that its radical 
axis with reference to a fixed circle always passes through 
a fixed point. Show that the locus of its centre is a circle 
having its centre at the fixed point. 

100. Show that the quadrilateral enclosed by the lines 
3x + 2y = 0, 2x - 3y + 1 = 0, 2x - Zy = 0, Ix + 2y = 1 
is a square. 



MISCELLANEOUS EXERCISES HI 

101. Show that the centre of the circle 

x 2 + y 2 - 2gx - 2/y + c + I (x 2 + y 2 - Ihx - 2ky + d) = 
divides the st. line joining the centres of the circles 
x i + y i _ 2gx - 2fy + c = and a; 2 + y 2 - 2/ia - Iky + d=0 
in the ratio of 1:1. 

102. The sum of the J_s from two fixed points (a^, y 1 ) 
and (.r.„ y„) to a variable line Ix + wy + n = is equal to 
the constant a. Prove that the line is always tangent to a 
fixed circle, and find the equation of the circle. (Problems — 
1907.) 

103. Show that the circles x 2 + y 2 + 2x - 8y + 8 = 0, 
x 2 + y 2 + Wx - 2y + 22 =0 touch each other. (Prob- 
lems— 1911.) 

104. Through one angular point A of a square ABCD a 
st. line is drawn meeting the sides BC and DC produced 
at E and F respectively. If ED and FB intersect in G, 
show that CG is _L EF. (Problems— 1913.) 

105. Prove that the lines 

ax + (b + c) y = b 2 + be + c 2 , 
bx + (c + a) y = c 2 + ca + a 2 , 
ex + (a + b) y = a 2 + ab + b 2 t 

are concurrent, and find the coordinates of their common 

point. (Problems — 1913.". 

106. Two circles whose centres are C, C, touch each 
other, internally at O. A st. line OPP' is drawn cutting 
the circles at P and P'. Show that the locus of the 
intersection of CP' and C'P is a circle whose diameter is a 
harmonic mean between the radii of the given circles ; and 
whose centre is at C" on the line OCC such that OC" is 
the harmonic mean between OC and OC. (Problems — 1913.) 

107. Prove that the chords of intersection with a fixed 
circle of all circles through two fixed points are concurrent. 
(Problems— 1912. ) 



112 ELEMENTARY ANALYTICAL GEOMETRY 

108. Find the equation of two st. lines through the origin 
and such that the J_s to them from the point (A, k) are 

+ d and - d. 

109. If axes of reference are drawn on a sheet of paper 
and if this is folded about the line joining (1, 3) to (2, 0), 
find the coordinates of the point which falls on (x, y). 

Find also the equation of the circle which coincides with 
x- + y 2 - 2y = 4. (Problems— 1917.) 

110. AS and AT, BP and BQ are tangents from any two 
points A and B to a fixed circle. C, D, E, F are the middle 
points of AS, AT, BP, BQ respectively. Prove that CD and 
EF, produced if necessary, meet on the line that bisects 
AB at rt. l s. (Problems— 1907.) 



ANSWERS 



§6. (Page 5.) 
4. (0, 0), (2a, 0;, (a, a,/3). 5. (0, 0), (6, 0), (6, b), (0, 6). 





§16. (Page 11.) 




4. 
5. 


1304. 12 . $^65. 
10, 17, 9. 13 . 3 or _ 13 . 




6. 

7. 

8. 
9. 

11. 


(a) t/58, j/82, 10 ; 14. 4* - 10j/ + 29 = 0. 

(6) ^v/2347 3/57 £i/30tT i5 - (W, W) ! 3 '7 nearly. 

•mT 2/107 v/857 •»! 1G - < 7 > 2 >' < - W W- 

7v/27 /137. 20. (*+*+»* *+%+») 

(0, 0)-the origin. „ ,~ -,. 
(t,|)and(^,f). 22 ; (li8f_ 14) . 




§19. (Page 16.) 




3. 


36-5. 8. 36 miles. 




4. 


25 5. 9. 1:3. 





§22. (Page 21.) 

2. y = bx. 6. 2ari-y+5=0t 

3. (o) The axis of x ; (b) The 7. a: 2 + y 2 - 8u; - 6j/ = 0. 

axis of y. 8. 5c 2 + i/ 2 -2^ + 4y = 31. 

4. x=4. 10. a;-2!/ + 3 = 0. 

5. 3x-oy = 17. 

§24. (Page 23.) 

1. (a) (2, 7) ; (6) (3, 2) ; (c) 3. (0, 0) and (6, 0). 

(2, -3); (d) (8, -6) and 4. 2x = a. 

(-8, 6); (e) (5, 12) and 5. 7*+4y=20. 

(-W it)- °- »: 2 + y 2 =9. 

2. (0, 9) and ( - 15, 0). 



114 ANSWERS 

§27. (Page 27.) 

_ . N a v ., ... x y 8. Sides, x + 9y + 14 = 0; 5x + 

2- (a) T + j-l;<6) r + t= 3y + 28 = 0, 2x-3y-14 

' v ' 3 8 Medians, x - 5y = 14, x + 2y 

3. (a) x - 3y = ; (6) 9x - 2y+ = ~7, 3a: -y=0 ; 

13 = ; (c) 8* + Hi/ + 34 Centroid ( - 1, - 3). 

= 9. Sides, 2x-5y=-24, 3x + 

Intercepts :-(a) 0, 0; (6) 2 ^ = 2 ' *"y = 4 ' 3x + 4 ^ 

_ J? 13 . / c \ _ 34 _ 34 — 33 ; 

5 ' J ' w 8 ' IT " Diagonals, 8x-y = 18, x + 

4 - ^~ 5 ' ~ 3 ^- 9y = 34,87x + 438y = 5948; 

"' ' ^' ■*-2t/' Line through middle points 

6- m, 4). of diagonals, 2x = 5. 

7. Ratio of equality. 10. ( - 6, - 7), (5, - 2), ( - 3, 4). 

§40. (Page 38.) 

3. («) x = i/3y ; (6) y + x v / 3 = 0; 8. (a) C=0 ; (6) A = ; (c) B 

(c)5x-7y + 35 = 0; (d) 3y =0; (rf) A = B; (e) A + B 

±4x + 9 = 0; (e)x-y = l. =0. 

4. (a) 9, -i -^);(6)(|,0). 9. m = l. 

5. (3, shy 10 - m =-i J= 2 ?- 



6. x + y + l = 0. 

7. vf\ i/58 ; tan ~ l |. 


11. 

13. 


« = 8, 6 = -4. 

C'-C 

Intercept— R 


§44. 


(Page 


42.) 


1. (a) 45° ; (6) 30° ; (c) 90° 

(d) tan _1 5. 
3. 9x + 4y + 47 = 0. 
5. 2x + 3y = 14. 


; 11. 

12. 
14. 


12 :5. 

(17 v/ 3 - 16)x + 47y = 344 


6. 7x + 5y = 4. 

7. 5x-3y + 8 = 0. 




+ 34 v /3"; and(17i/iW 26) 
x-47y = 34v/3"-344. 


8. x — y + 2 = and x + y - 

12 = 0. 

9. a-y'3y = 3i/3 _ -5 and x4 

v /3~y=-3v / 3~-5. 
10. (2f, -4}). 


15. 

17. 
18. 


6x + y + 8 = : and x - 6y 
• =11. 

/a 6*+c*-a6\ 

\ 2' 2c / 

Bx-Ay+j9i/A 2 +B 2 = 0. 



ANSWERS 



115 



§54. (Page 53.) 



38 24 12 T /2 

(a)— -;( 6 )_ ;(c) -— - 

• 29 x /-" 
5 



!/13 

(<0 



35 
•71" 



C2-C1 

•a 2 +6* 

( - 24, 55). 
8x+6y=15. 

x-y = 0. 
2ft- 

(wii - m) (ax - 61/) + ^(rnjC - 

mc l ) + a(c 1 -c) = 0. 
abc - ap - bg* - c/r + 2fgh = 0. 
(6 - rrija - c x ) (1/ - mx - c) = 

(6 - ma — c) (y — rrijx — c x ). 
(AC, - A!C)x+ (BCi - BiC)!/ 

= 0. 

• 2. 
±2/130-11 

7 
x + 21y = 6 and 189x-9i/ = 

692. 
3x + y = 2 and x-3y = 24. 



19. 

20. 

21. 
22. 
24. 

25. 
26. 

27. 

28. 
29. 



30. 

32. 
33. 
34. 
35. 



(Mi, Ml)- 
33x + 61y=216. 

(7 •3 + 11) 



11 

§•* 

a& (c - 



13 



d) 



•a 2 + 6 2 
15x + 14i/ = 100. 
13x-7y = 3 and 7x + 13y = 
119. 

(24 + 13 • 3) x + 23*/ + 52/3 
+ 257=0and(24-13/3> 

+ 23j/-52j/3 + 257 = 0. 
ox - 12y + 56 = and 5x - 

12;/ -74 = 0. 
6x + v = 31, x-y + 3 = 0; 
14 7 

•37' >/2" 

(51*. 3ft)- 

12x-5j/ = 26, y=2. 

Ax + B y +p •A 2 +B s = 0. 
Bx-Ay+Afc-B7i=0. 

(•)(W.-A);(6)(-lft 



§ 60. (Page fin.) 



(a) x = a, x = b ; 

(6) x-i/ = 0, x+y=0; 

(c) x = 0, x = 3i/; 

(d) 2x-y=0, 4x-3y=0; 

(e) x = a, 1/= -6 ; 
(/)3x-y=-4, x-3y = 5. 



2. $ and £. 

3. The axes of coordinates. 

4. (d) ton-i ft ; (e) 90° ; (J) 

tan'* |. 

5. bc = ad. 

6. 8. 

7. 3x 2 -y 2 -30x + 6y + 66 = 0. 



116 ANSWERS 

§65. (Page 64.) 

1. 2x l -llxy + 12i/ 2 = 0. 6. 2xy + a 2 = 0. 

2. x*+xy=0. 8. 9«'-25i/ , =0. 

3. (-|, |); 2(x 2 + y 2 ) = 19. 9. tan' 1 | ; 3x 2 + 2y 2 = 10. 
5. (A/r+B)x + (B l /F-A)y+ 

2C = 0. 

§ 66. (Page 66.) 

1. (a) p/145; 33. ( 4 iV "tW- _ 

/h\ / ~ ixiA ; , v-tn 34 - 2j/-7x±xV55=0. 
(6) j/oa^ + lbab + 136 z ; 

(c) a + 6. 35. 

2- P(5, 3|); Q(17£, 16£). V153 

3. fab. 3 7- V21 

4. 41i ' 2 

5. 10|. 38. x + 6y = 39and6x-y = 12. 

6. Xi (!/2 - y a ) + ^ (■{/,- t/i) + 41. 2x + 3y + 11 = 0. 

^(!/i-J/ 2 ) = 0. 42. I65x + 605y + 3114 = 0;332x 

7. 8x + 7y = 5. -747y + 3466 = ; 497x 
9. 2ax + 26y = c + d\ -142y- 2178 = 0. 

10. x=y ia»a+6. 43. (§,§). 

13. xcosa + i/swia=a. 44. (a) (ah-cf)x + (bh-cg)y = 0; 

14. fan- 1 W- ( 6 > ^ ~ V) < x ~ ^ + c ^ + ») 



15. m = f, 6 = -4J. 



/i(a + 6) = 0. 



16. x + i/ v /3 + 2 U /3-l) = 0. 45. 5x-14y=6 and 154x + 55y 

17< ^mfe-fe + a ^ 46 49x _ 2 45y=242. 



sm a - m e<j.s a 



48. x + y = 10. 



18. 5 + l = A + A. 49. x-y = 10, ory-x = 10. 

rt fc a 6 50. 7x + 5y + 50 = 0, and 9x + 35y 

20. 6x + ay=ad. +150 = 0. 

21. (-2, -8) and (-6, -14). 51 K _ 4 ^a =1 i. 

23. a = 6 or - 4 ; (J, $), (£, - i). 52 g( a - h)x + 2(6 - % = a 2 + 6 2 

25. 45°. -h 2 -k\ 

26. 65x-65y + 14a + 36 = 0. 53. 2x + y + 15 = 0. 

27. 6(am + c)x + (6d - a6 + adm 54. 45°. 

+ ac)y - bd(am + c) = 0. 55 1Qx + 4,, + n = , and 4* - 
29 ab 10(/-*33 = 0. 

VoMT*" 56. x 2 -y 2 = 0. 

30. x + 4y + l = 0. 57. -13orl0. 1 ; . 

31. x+7y+6=0. 59. (11, 1), (-1, -5),or(-6,7> 

32. (6^V, 5}|f)i (-3 7 V, ff); 61. - |J. 

(26 T 2 r <V, 4^) ; (0, 57). 62. 24. 



ANSWERS 117 


§72. (Page 75.) 


1. x 2 + y t = 5. 


13. x 2 + y 2 -4x + 6y = 16. 


2. (x-ey+( y -2y=9. 


15. 47 (x 2 + y 2 ) - 181x + 341 y - 


3. (x + 5)' + (y + l) 2 = 26. 


1996 = 0. 


4. a 2 + b 2 = c 2 . 


16. y'lO. 


5. (a) (3^1), 5; (&)(-£, -f), 


17. 14 T /2T 


^j (c) (7, 0), 7; (d) 


21. 2x-5i/ = 15. 


o 


22. 123. 


(0, -b), VV + c\ 


23. 14(x+y)=17. 


8. x 2 + y'-x-7!/ = 0. 


24. c = c'. 


9. x*+y 2 -3x=19. 


25. (h + fe) (x 2 + y 2 ) - (h 2 + W) (x + 


11. (a)/=0; (6) = 0. 


y) = 0. 


12. x 2 + y 2 -4x + 4</ = 2. 




§81. (Page 85.) 


1. (a) 3x + 5y = 34; (b) 3x-4y 
+ 11 = 0; (c)12x + 6y= 


16. x-3y-3; (^±3/119 

y V io 


96; (d) & x+/j/ = 0. 
2. y=x ± j/70. 


-st^/im 
10 / 


3. (o)Ax+By=±r>/A ! +B :f ; 


17. (A' + B 2 )(x 2 + y 2 ) = C 2 . 


(6) Bx - Ay - 


18. 34 (x 2 + y 2 ) -476x - 136y + 


±r,/A* + B 2 . 


1753 = 0. 


4. (2, 4). 


19. (1, I); (17, -13). 


5. (6, 1). 


transverse, 12(5y - 7) = ( — 21 


6. (a)C* = r»(A* + B 2 ); (6)(Agr 


±5/15) (5x-l); 


+ B/ - C) 2 = (A 2 + B 2 ) 


direct, 24(y + 13) = (-21± 


('f+f 2 -c). 


!/21)(x-17). 


7. a 2 6 2 = r 2 (a 2 + 6 2 ). 


20. direct, y = 9 and 3x + 4y + 3 
= 0; 


8. c = 9 J . 

9. x 2 + y 2 -2(7±2/5)(x + y) + 

69128/5 = 0. 


transverse are imaginary. 
(2, 9)and(-l, 0) on x 2 + y 2 
-4x-8y-5 = 0; 


10. x- v /3'y + 3 l /3"-l = 0, and 


(5, 9) and (|, -f)on x 2 + y 2 


x-v/3y-5,/3~-l = 0. 


-10x-6i/-2 = 0. 


11. VW- 


21. x + 3y + g + 3f± 


13. (x - g) cos a + ((/-/) sin « = r. 


/10(^+/ 2 -c) = 0. 


14- (V,-¥); 4x + 3y + l = 0. 


22. x-j/3y±10 = 0. 



118 ANSWERS 

§92. (Page 94.) 

1. (a) 3»+5y=30 ; (6) ax=r 2 ; 4 /c*_ c 2 \ 

(c)4x + 17 = 0; (cQlOx- ' '«' 6 / 

2i/ = 7; (e) ^x + /cj/ = /i 2 + _ /fn-g^lc-fgm 

fc*-r*. °- V l+/ m + ^ ' 

2. (a) (2, -7) ; (b){ -U, V) ; g'm-f-mc~fgl \ 

(c) ( - 35, 11); (d) (0, 0). 1 +/m + g Z / 

3. (c) •Sx-4y + 25 = 0; (e) 9x 6. (1, 4). 

+ 13./ = 25; (/) 13x - 9y 
= 0; ( ? ) 24x - 7y = 125. 

§ 95. (Page 97.) 

1. (a) G ; (b) 5 ; (c) 8 ; (cZ) T /<T 3. x 2 + i/ 2 -16x + 51=0. 

2. x' + i/ 2 -10x-4i/ = 7. 4. 4x + 3y = 25and3x-4i/ = 26. 





§98. 


(Page 


9S.) 


1. 12x + 8y = 95. 






4. 


(-13, -7> 


2. 9x-8i/ + 15 = 0. 










MlSCELI^ 


LNEOUS 


Exercises. (Page 99. '< 


1. 3x + 4y=25. 






16. 


x 2 + i/-Gx-3i/ = 0. 


2. 2x-lli/ + 329 = 0. 






17. 


3x-y = 3k±kjTd. 


3. 113. 






18. 


/"3 8 2 3 0Q\ 


4. 29. 






20. 


x 2 + y 2 = Tix. 


o. — + — = 2. 






21. 


3=2 + V 1 — ax - by = 0. 


/i fc 






22. 


x 2 +if-gx-fy = 0. 


6. (a) = -; 

Xj-x.2 i/i — y a 


(b)( Xl 


- 


23. 
24. 


(if, fi). 

8(a; 2 + ,/; _ 25x + 75y+71 


^(x-^+d/i- 


1/2)(l/- 


k) 




= 0. 


= 0. 






25. 


lOorf. 


8. 7 : 4. 






L'o. 


Cencre divides AB exteis 


v - MIS' 115/- 








nally in ratio A; 2 : 1. 


10. 77. 






27. 


(a) (7, 1); (6)AD = DB. 


12. x 2 + y 2 =12. 






30. 


(2, 1). 


13. (a + c, b + d). 






31. 


23 8 
1?5- 


15. The point (-a cos a, - 


a 


32. 


x' i + y' i -2h(x~a)-2k(y-b) 


sin a). 








= a 2 + 6 2 . 







ANSWERS 




Miscellaneous Exercises — Continued. 


33. 
35. 


x* + y 2 +2gx + 2fy + 2c- 
2x-(b + c)y + 2abc = 0. 


-?- 81. Ci ' b \ 
2m 

85. y 2 -x 2 = 0. 


37. 

40. 
41. 


^13 
4' 
19^-60x1/+ 44 »/ 2 = 0. 
3x 2 -8a;)/-3i/ 2 = 0. 


86. x 2 + y 2 -2a(x + y) + 
88. x 2 + y 2 = 2(x + y). 

92. 59(ar + y 2 ) - 44 (a; 
740. 



119 



102 . (.-ajs)' + 

(v-*?)'=T 



.„ , - 1 8 N /243 

42. tan — ^-. 

43. y=2 and 15a; + 8y = 31. 
46. (a + c-g, b+d-h). 
49. 163a; + 9y + 54 = and 239sc 1f)r - bc + ca + ab 

-573i/ + 1112 = 0. 1U °- a + b + c ' 

67. a; 2 + i/ 2 -(/ix + Ai/) = 0. a« + b' + c ' + 6c + ca + a6 

69. a: 2 + y s -10x + 9 = 0. a + b + c 

74 "* W 108. (/: 2 - c*V - 2^-xy + (h 2 - 

' 20* d 2 )y 2 = 0. 

76. The circle 3(x 2 + y 2 ) + 8x + 109. 18 - 4x - 3y , 6 - 3x + 4y, 

10*/ = 92. 5 5 

77. 3(x 2 + y 2 )-2a(x + y) = 2a t . x i +y 2 -6x-4y + 8 = Q. 



f 2r 



^3 J /3 # , v 

/L A Y. ^ - f^c'^.-^u, 

v 2- 






* O 



4-x, r- 


3 tf'^ y 








o 




" 3 J * * ' 

















a?